RD Sharma Solutions for Class 8 Chapter 4 Cubes and Cube Roots Free Online
EXERCISE 4.1 PAGE NO: 4.7
1. Find the cubes of the following numbers:
(i) 7 (ii) 12
(i) 7 (ii) 12
(iii) 16 (iv) 21
(v) 40 (vi) 55
(vii) 100 (viii) 302
(ix) 301
Solution:
(i) 7
Cube of 7 is
7 = 7× 7 × 7 = 343
(ii) 12
Cube of 12 is
12 = 12× 12× 12 = 1728
(iii) 16
Cube of 16 is
16 = 16× 16× 16 = 4096
(iv) 21
Cube of 21 is
21 = 21 × 21 × 21 = 9261
(v) 40
Cube of 40 is
40 = 40× 40× 40 = 64000
(vi) 55
Cube of 55 is
55 = 55× 55× 55 = 166375
(vii) 100
Cube of 100 is
100 = 100× 100× 100 = 1000000
(viii) 302
Cube of 302 is
302 = 302× 302× 302 = 27543608
(ix) 301
Cube of 301 is
301 = 301× 301× 301 = 27270901
2.Write the cubes of all natural numbers between 1 and 10 and verify the following statements:
(i) Cubes of all odd natural numbers are odd.
(ii) Cubes of all even natural numbers are even.
(i) Cubes of all odd natural numbers are odd.
(ii) Cubes of all even natural numbers are even.
Solutions:
Firstly let us find the Cube of natural numbers up to 10
13 = 1 × 1 × 1 = 1
23 = 2 × 2 × 2 = 8
33 = 3 × 3 × 3 = 27
43 = 4 × 4 × 4 = 64
53 = 5 × 5 × 5 = 125
63 = 6 × 6 × 6 = 216
73 = 7 × 7 × 7 = 343
83 = 8 × 8 × 8 = 512
93 = 9 × 9 × 9 = 729
103 = 10 × 10 × 10 = 1000
∴ From the above results we can say that
(i) Cubes of all odd natural numbers are odd.
(ii) Cubes of all even natural numbers are even.
3. Observe the following pattern:
13 = 1
13 = 1
13 + 23 = (1+2)2
13 + 23 + 33 = (1+2+3)2
Write the next three rows and calculate the value of 13 + 23 + 33 +…+ 93 by the above pattern.
Write the next three rows and calculate the value of 13 + 23 + 33 +…+ 93 by the above pattern.
Solution:
According to given pattern,
13 + 23 + 33 +…+ 93
13 + 23 + 33 +…+ n3 = (1+2+3+…+n) 2
So when n = 10
13 + 23 + 33 +…+ 93 + 103 = (1+2+3+…+10) 2
= (55)2 = 55×55 = 3025
4. Write the cubes of 5 natural numbers which are multiples of 3 and verify the followings:
“The cube of a natural number which is a multiple of 3 is a multiple of 27’
“The cube of a natural number which is a multiple of 3 is a multiple of 27’
Solution:
We know that the first 5 natural numbers which are multiple of 3 are 3, 6, 9, 12 and 15
So now, let us find the cube of 3, 6, 9, 12 and 15
33 = 3 × 3 × 3 = 27
63 = 6 × 6 × 6 = 216
93 = 9 × 9 × 9 = 729
123 = 12 × 12 × 12 = 1728
153 = 15 × 15 × 15 = 3375
We found that all the cubes are divisible by 27
∴ “The cube of a natural number which is a multiple of 3 is a multiple of 27’
5.Write the cubes of 5 natural numbers which are of the form 3n + 1 (e.g. 4, 7, 10, …) and verify the following:
“The cube of a natural number of the form 3n+1 is a natural number of the same from i.e. when divided by 3 it leaves the remainder 1’
“The cube of a natural number of the form 3n+1 is a natural number of the same from i.e. when divided by 3 it leaves the remainder 1’
Solution:
We know that the first 5 natural numbers in the form of (3n + 1) are 4, 7, 10, 13 and 16
So now, let us find the cube of 4, 7, 10, 13 and 16
43 = 4 × 4 × 4 = 64
73 = 7 × 7 × 7 = 343
103 = 10 × 10 × 10 = 1000
133 = 13 × 13 × 13 = 2197
163 = 16 × 16 × 16 = 4096
We found that all these cubeswhen divided by ‘3’ leaves remainder 1.
∴ the statement “The cube of a natural number of the form 3n+1 is a natural number of the same from i.e. when divided by 3 it leaves the remainder 1’ is true.
6. Write the cubes 5 natural numbers of the from 3n+2(i.e.5,8,11….) and verify the following:
“The cube of a natural number of the form 3n+2 is a natural number of the same form i.e. when it is dividend by 3 the remainder is 2’
“The cube of a natural number of the form 3n+2 is a natural number of the same form i.e. when it is dividend by 3 the remainder is 2’
Solution:
We know that the first 5 natural numbers in the form (3n + 2) are 5, 8, 11, 14 and 17
So now, let us find the cubes of 5, 8, 11, 14 and 17
53 = 5 × 5 × 5 = 125
83 = 8 × 8 × 8 = 512
113 = 11 × 11 × 11 = 1331
143 = 14 × 14 × 14 = 2744
173 = 17 × 17 × 17 = 4313
We found that all these cubes when divided by ‘3’ leaves remainder 2.
∴ thestatement“The cube of a natural number of the form 3n+2 is a natural number of the same form i.e. when it is dividend by 3 the remainder is 2’ is true.
7.Write the cubes of 5 natural numbers of which are multiples of 7 and verity the following:
“The cube of a multiple of 7 is a multiple of 73.
“The cube of a multiple of 7 is a multiple of 73.
Solution:
The first 5 natural numbers which are multiple of 7 are 7, 14, 21, 28 and 35
So, the Cube of 7, 14, 21, 28 and 35
73 = 7 × 7 × 7 = 343
143 = 14 × 14 × 14 = 2744
213 = 21× 21× 21 = 9261
283 = 28 × 28 × 28 = 21952
353 = 35 × 35 × 35 = 42875
We found that all these cubes are multiples of 73(343) as well.
∴The statement“The cube of a multiple of 7 is a multiple of 73 is true.
8. Which of the following are perfect cubes?
(i) 64 (ii) 216
(iii) 243 (iv) 1000
(v) 1728 (vi) 3087
(vii) 4608 (viii) 106480
(ix) 166375 (x) 456533
(i) 64 (ii) 216
(iii) 243 (iv) 1000
(v) 1728 (vi) 3087
(vii) 4608 (viii) 106480
(ix) 166375 (x) 456533
Solution:
(i) 64
First find the factors of 64
64 = 2 × 2 × 2 × 2 × 2 × 2 = 26 = (22)3 = 43
Hence, it’s a perfect cube.
(ii) 216
First find thefactors of 216
216 = 2 × 2 × 2 × 3 × 3 × 3 = 23 × 33 = 63
Hence, it’s a perfect cube.
(iii) 243
First find thefactors of 243
243 = 3 × 3 × 3 × 3 × 3 = 35 = 33 × 32
Hence, it’s not a perfect cube.
(iv) 1000
First find thefactors of 1000
1000 = 2 × 2 × 2 × 5 × 5 × 5 = 23 × 53 = 103
Hence, it’s a perfect cube.
(v) 1728
First find thefactors of 1728
1728 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 = 26 × 33 = (4 × 3 )3 = 123
Hence, it’s a perfect cube.
(vi) 3087
First find thefactors of 3087
3087 = 3 × 3 × 7 × 7 × 7 = 32 × 73
Hence, it’s not a perfect cube.
(vii) 4608
First find thefactors of 4608
4608 = 2 × 2 × 3 × 113
Hence, it’s not a perfect cube.
(viii) 106480
First find thefactors of 106480
106480 = 2 × 2 × 2 × 2 × 5 × 11 × 11 × 11
Hence, it’s not a perfect cube.
(ix) 166375
First find thefactors of 166375
166375= 5 × 5 × 5 × 11 × 11 × 11 = 53 × 113 = 553
Hence, it’s a perfect cube.
(x) 456533
First find thefactors of 456533
456533= 11 × 11 × 11 × 7 × 7 × 7 = 113 × 73 = 773
Hence, it’s a perfect cube.
9. Which of the following are cubes of even natural numbers?
216, 512, 729, 1000, 3375, 13824
216, 512, 729, 1000, 3375, 13824
Solution:
(i) 216 = 23 × 33 = 63
It’s a cube of even natural number.
(ii) 512 = 29 = (23)3 = 83
It’s a cube of even natural number.
(iii) 729 = 33 × 33 = 93
It’s not a cube of even natural number.
(iv) 1000 = 103
It’s a cube of even natural number.
(v) 3375 = 33 × 53 = 153
It’s not a cube of even natural number.
(vi) 13824 = 29 × 33 = (23)3 × 33 = 83×33 = 243
It’s a cube of even natural number.
10. Which of the following are cubes of odd natural numbers?
125, 343, 1728, 4096, 32768, 6859
125, 343, 1728, 4096, 32768, 6859
Solution:
(i) 125 = 5 × 5 × 5 × 5 = 53
It’s a cube of odd natural number.
(ii) 343 = 7 × 7 × 7 = 73
It’s a cube of odd natural number.
(iii) 1728 = 26 × 33 = 43 × 33 = 123
It’s not a cube of odd natural number. As 12 is even number.
(iv) 4096 = 212 = (26)2 = 642
Its not a cube of odd natural number. As 64 is an even number.
(v) 32768 = 215 = (25)3 = 323
It’s not a cube of odd natural number. As 32 is an even number.
(vi) 6859 = 19 × 19 × 19 = 193
It’s a cube of odd natural number.
11. What is the smallest number by which the following numbers must be multiplied, so that the products are perfect cubes?
(i) 675 (ii) 1323
(iii) 2560 (iv) 7803
(v) 107811 (vi) 35721
(i) 675 (ii) 1323
(iii) 2560 (iv) 7803
(v) 107811 (vi) 35721
Solution:
(i) 675
First find the factors of 675
675 = 3 × 3 × 3 × 5 × 5
= 33 × 52
∴To make a perfect cube we need to multiply the product by 5.
(ii) 1323
First find the factors of 1323
1323 = 3 × 3 × 3 × 7 × 7
= 33 × 72
∴To make a perfect cube we need to multiply the product by 7.
(iii) 2560
First find the factors of 2560
2560 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 5
= 23 × 23 × 23 × 5
∴To make a perfect cube we need to multiply the product by 5 × 5 = 25.
(iv) 7803
First find the factors of 7803
7803 = 3 × 3 × 3 × 17 × 17
= 33 × 172
∴To make a perfect cube we need to multiply the product by 17.
(v) 107811
First find the factors of 107811
107811 = 3 × 3 × 3 × 3 × 11 × 11 × 11
= 33 × 3 × 113
∴To make a perfect cube we need to multiply the product by 3 × 3 = 9.
(vi) 35721
First find the factors of 35721
35721 = 3 × 3 × 3 × 3 × 3 × 3 × 7 × 7
= 33 × 33 × 72
∴To make a perfect cube we need to multiply the product by 7.
12. By which smallest number must the following numbers be divided so that the quotient is a perfect cube?
(i) 675 (ii) 8640
(iii) 1600 (iv) 8788
(v) 7803 (vi) 107811
(vii) 35721 (viii) 243000
(i) 675 (ii) 8640
(iii) 1600 (iv) 8788
(v) 7803 (vi) 107811
(vii) 35721 (viii) 243000
Solution:
(i) 675
First find the prime factors of 675
675 = 3 × 3 × 3 × 5 × 5
= 33 × 52
Since 675 is not a perfect cube.
To make the quotient a perfect cube we divide it by 52 = 25, which gives 27 as quotient where, 27 is a perfect cube.
∴ 25 is the required smallest number.
(ii) 8640
First find the prime factors of 8640
8640 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 5
= 23 × 23 × 33 × 5
Since 8640 is not a perfect cube.
To make the quotient a perfect cube we divide it by 5, which gives 1728 as quotient and we know that 1728 is a perfect cube.
∴5 is the required smallest number.
(iii) 1600
First find the prime factors of 1600
1600 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5
= 23 × 23 × 52
Since 1600 is not a perfect cube.
To make the quotient a perfect cube we divide it by 52 = 25, which gives 64 as quotient and we know that 64 is a perfect cube
∴ 25 is the required smallest number.
(iv) 8788
First find the prime factors of 8788
8788 = 2 × 2 × 13 × 13 × 13
= 22 × 133
Since 8788 is not a perfect cube.
To make the quotient a perfect cube we divide it by 4, which gives 2197 as quotient and we know that 2197 is a perfect cube
∴ 4 is the required smallest number.
(v) 7803
First find the prime factors of 7803
7803 = 3 × 3 × 3 × 17 × 17
= 33 × 172
Since 7803 is not a perfect cube.
To make the quotient a perfect cube we divide it by 172 = 289, which gives 27 as quotient and we know that 27 is a perfect cube
∴ 289 is the required smallest number.
(vi) 107811
First find the prime factors of 107811
107811 = 3 × 3 × 3 × 3 × 11 × 11 × 11
= 33 × 113 × 3
Since 107811 is not a perfect cube.
To make the quotient a perfect cube we divide it by 3, which gives 35937 as quotient and we know that 35937 is a perfect cube.
∴ 3 is the required smallest number.
(vii) 35721
First find the prime factors of 35721
35721 = 3 × 3 × 3 × 3 × 3 × 3 × 7 × 7
= 33 × 33 × 72
Since 35721 is not a perfect cube.
To make the quotient a perfect cube we divide it by 72 = 49, which gives 729 as quotient and we know that 729 is a perfect cube
∴ 49 is the required smallest number.
(viii) 243000
First find the prime factors of 243000
243000 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 5 × 5 × 5
= 23 × 33 × 53 × 32
Since 243000 is not a perfect cube.
To make the quotient a perfect cube we divide it by 32 = 9, which gives 27000 as quotient and we know that 27000 is a perfect cube
∴ 9 is the required smallest number.
13. Prove that if a number is trebled then its cube is 27 time the cube of the given number.
Solution:
Let us consider a number as a
So the cube of the assumed number is = a3
Now, the number is trebled = 3 × a = 3a
So the cube of new number = (3a) 3 = 27a3
∴New cube is 27 times of the original cube.
Hence, proved.
14. What happens to the cube of a number if the number is multiplied by
(i) 3?
(ii) 4?
(iii) 5?
(i) 3?
(ii) 4?
(iii) 5?
Solution:
(i) 3?
Let us consider the number as a
So its cube will be = a3
According to the question, the number is multiplied by 3
New number becomes = 3a
So the cube of new number will be = (3a) 3 = 27a3
Hence, number will become 27 times the cube of the number.
(ii) 4?
Let us consider the number as a
So its cube will be = a3
According to the question, the number is multiplied by 4
New number becomes = 4a
So the cube of new number will be = (4a) 3 = 64a3
Hence, number will become 64 times the cube of the number.
(iii) 5?
Let us consider the number as a
So its cube will be = a3
According to the question, the number is multiplied by 5
New number becomes = 5a
So the cube of new number will be = (5a) 3 = 125a3
Hence, number will become 125 times the cube of the number.
15. Find the volume of a cube, one face of which has an area of 64m2.
Solution:
We know that the given area of one face of cube = 64 m2
Let the length of edge of cube be ‘a’ metre
a2 = 64
a = √ 64
= 8m
Now, volume of cube = a3
a3 = 83 = 8 × 8 × 8
= 512m3
∴Volume of a cube is 512m3
16. Find the volume of a cube whose surface area is 384m2.
Solution:
We know that the surface area of cube = 384 m2
Let us consider the length of each edge of cube be ‘a’ meter
6a2 = 384
a2 = 384/6
= 64
a = √64
= 8m
Now, volume of cube = a3
a3 = 83 = 8 × 8 × 8
= 512m3
∴ Volume of a cube is 512m3
17. Evaluate the following:
(i) {(52 + 122)1/2}3
(ii) {(62 + 82)1/2}3
(i) {(52 + 122)1/2}3
(ii) {(62 + 82)1/2}3
Solution:
(i) {(52 + 122)1/2}3
When simplified above equation we get,
{(25 + 144)1/2}3
{(169)1/2}3
{(132)1/2}3
(13)3
2197
(ii) {(62 + 82)1/2}3
When simplified above equation we get,
{(36 + 64)1/2}3
{(100)1/2}3
{(102)1/2}3
(10)3
1000
18. Write the units digit of the cube of each of the following numbers:
31, 109, 388, 4276, 5922, 77774, 44447, 125125125
31, 109, 388, 4276, 5922, 77774, 44447, 125125125
Solution:
31
To find unit digit of cube of a number we perform the cube of unit digit only.
Unit digit of 31 is 1
Cube of 1 = 13 = 1
∴ Unit digit of cube of 31 is always 1
109
To find unit digit of cube of a number we perform the cube of unit digit only.
Unit digit of 109 is = 9
Cube of 9 = 93 = 729
∴ Unit digit of cube of 109 is always 9
388
To find unit digit of cube of a number we perform the cube of unit digit only.
Unit digit of 388 is = 8
Cube of 8 = 83 = 512
∴ Unit digit of cube of 388 is always 2
4276
To find unit digit of cube of a number we perform the cube of unit digit only.
Unit digit of 4276 is = 6
Cube of 6 = 63 = 216
∴ Unit digit of cube of 4276 is always 6
5922
To find unit digit of cube of a number we perform the cube of unit digit only.
Unit digit of 5922 is = 2
Cube of 2 = 23 = 8
∴ Unit digit of cube of 5922 is always 8
77774
To find unit digit of cube of a number we perform the cube of unit digit only.
Unit digit of 77774 is = 4
Cube of 4 = 43 = 64
∴ Unit digit of cube of 77774 is always 4
44447
To find unit digit of cube of a number we perform the cube of unit digit only.
Unit digit of 44447 is = 7
Cube of 7 = 73 = 343
∴ Unit digit of cube of 44447 is always 3
125125125
To find unit digit of cube of a number we perform the cube of unit digit only.
Unit digit of 125125125 is = 5
Cube of 5 = 53 = 125
∴ Unit digit of cube of 125125125 is always 5
19. Find the cubes of the following numbers by column method:
(i) 35
(ii) 56
(iii) 72
(i) 35
(ii) 56
(iii) 72
Solution:
(i) 35
We have, a = 3 and b = 5
Column I
a3
 Column II
3×a2×b
 Column III
3×a×b2
 Column IV
b3

33 = 27  3×9×5 = 135  3×3×25 = 225  53 = 125 
+15  +23  +12  125 
42  158  237  
42  8  7  5 
∴ The cube of 35 is 42875
(ii) 56
We have, a = 5 and b = 6
Column I
a3
 Column II
3×a2×b
 Column III
3×a×b2
 Column IV
b3

53 = 125  3×25×6 = 450  3×5×36 = 540  63 = 216 
+50  +56  +21  126 
175  506  561  
175  6  1  6 
∴ The cube of 56 is 175616
(iii) 72
We have, a = 7 and b = 2
Column I
a3
 Column II
3×a2×b
 Column III
3×a×b2
 Column IV
b3

73 = 343  3×49×2 = 294  3×7×4 = 84  23 = 8 
+30  +8  +0  8 
373  302  84  
373  2  4  8 
∴ The cube of 72 is 373248
20. Which of the following numbers are not perfect cubes?
(i) 64
(ii) 216
(iii) 243
(iv) 1728
(i) 64
(ii) 216
(iii) 243
(iv) 1728
Solution:
(i) 64
Firstly let us find the prime factors of 64
64 = 2 × 2 × 2 × 2 × 2 × 2
= 23 × 23
= 43
Hence, it’s a perfect cube.
(ii) 216
Firstly let us find the prime factors of 216
216 = 2 × 2 × 2 × 3 × 3 × 3
= 23 × 33
= 63
Hence, it’s a perfect cube.
(iii) 243
Firstly let us find the prime factors of 243
243 = 3 × 3 × 3 × 3 × 3
= 33 × 32
Hence, it’s not a perfect cube.
(iv) 1728
Firstly let us find the prime factors of 1728
1728 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3
= 23 × 23 × 33
= 123
Hence, it’s a perfect cube.
21. For each of the nonperfect cubes in Q. No 20 find the smallest number by which it must be
(a) Multiplied so that the product is a perfect cube.
(b) Divided so that the quotient is a perfect cube.
(a) Multiplied so that the product is a perfect cube.
(b) Divided so that the quotient is a perfect cube.
Solution:
Only nonperfect cube in previous question was = 243
(a) Multiplied so that the product is a perfect cube.
Firstly let us find the prime factors of 243
243 = 3 × 3 × 3 × 3 × 3 = 33 × 32
Hence, to make it a perfect cube we should multiply it by 3.
(b) Divided so that the quotient is a perfect cube.
Firstly let us find the prime factors of 243
243 = 3 × 3 × 3 × 3 × 3 = 33 × 32
Hence, to make it a perfect cube we have to divide it by 9.
22. By taking three different, values of n verify the truth of the following statements:
(i) If n is even, then n3 is also even.
(ii) If n is odd, then n3 is also odd.
(ii) If n leaves remainder 1 when divided by 3, then n3 also leaves 1 as remainder when divided by 3.
(iv) If a natural number n is of the form 3p+2 then n3 also a number of the same type.
(i) If n is even, then n3 is also even.
(ii) If n is odd, then n3 is also odd.
(ii) If n leaves remainder 1 when divided by 3, then n3 also leaves 1 as remainder when divided by 3.
(iv) If a natural number n is of the form 3p+2 then n3 also a number of the same type.
Solution:
(i) If n is even, then n3 is also even.
Let us consider three even natural numbers 2, 4, 6
So now, Cubes of 2, 4 and 6 are
23 = 8
43 = 64
63 = 216
Hence, we can see that all cubes are even in nature.
Statement is verified.
(ii) If n is odd, then n3 is also odd.
Let us consider three odd natural numbers 3, 5, 7
So now, cubes of 3, 5 and 7 are
33 = 27
53 = 125
73 = 343
Hence, we can see that all cubes are odd in nature.
Statement is verified.
(iii) If n leaves remainder 1 when divided by 3, then n3 also leaves 1 as remainder when divided by 3.
Let us consider three natural numbers of the form (3n+1) are 4, 7 and 10
So now, cube of 4, 7, 10 are
43 = 64
73 = 343
103 = 1000
We can see that if we divide these numbers by 3, we get 1 as remainder in each case.
Hence, statement is verified.
(iv) If a natural number n is of the form 3p+2 then n3 also a number of the same type.
Let us consider three natural numbers of the form (3p+2) are 5, 8 and 11
So now, cube of 5, 8 and 10 are
53 = 125
83 = 512
113 = 1331
Now, we try to write these cubes in form of (3p + 2)
125 = 3 × 41 + 2
512 = 3 × 170 + 2
1331 = 3 × 443 + 2
Hence, statement is verified.
23. Write true (T) or false (F) for the following statements:
(i) 392 is a perfect cube.
(ii) 8640 is not a perfect cube.
(iii) No cube can end with exactly two zeros.
(iv) There is no perfect cube which ends in 4.
(v) For an integer a, a3 is always greater than a2.
(vi) If a and b are integers such that a2>b2, then a3>b3.
(vii) If a divides b, then a3 divides b3.
(viii) If a2 ends in 9, then a3 ends in 7.
(ix) If a2 ends in an even number of zeros, then a3 ends in 25.
(x) If a2 ends in an even number of zeros, then a3 ends in an odd number of zeros.
(i) 392 is a perfect cube.
(ii) 8640 is not a perfect cube.
(iii) No cube can end with exactly two zeros.
(iv) There is no perfect cube which ends in 4.
(v) For an integer a, a3 is always greater than a2.
(vi) If a and b are integers such that a2>b2, then a3>b3.
(vii) If a divides b, then a3 divides b3.
(viii) If a2 ends in 9, then a3 ends in 7.
(ix) If a2 ends in an even number of zeros, then a3 ends in 25.
(x) If a2 ends in an even number of zeros, then a3 ends in an odd number of zeros.
Solution:
(i) 392 is a perfect cube.
Firstly let’s find the prime factors of 392 = 2 × 2 × 2 × 7 × 7 = 23 × 72
Hence the statement is False.
(ii) 8640 is not a perfect cube.
Prime factors of 8640 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 5 = 23 × 23 × 33 × 5
Hence the statement is True
(iii) No cube can end with exactly two zeros.
Statement is True.
Because a perfect cube always have zeros in multiple of 3.
(iv) There is no perfect cube which ends in 4.
We know 64 is a perfect cube = 4 × 4 × 4 and it ends with 4.
Hence the statement is False.
(v) For an integer a, a3 is always greater than a2.
Statement is False.
Because in case of negative integers ,
(2)2 = 4 and (2)3 = 8
(vi) If a and b are integers such that a2>b2, then a3>b3.
Statement is False.
In case of negative integers,
(5)2 > (4)2 = 25 > 16
But, (5)3 > (4)3 = 125 > 64 is not true.
(vii) If a divides b, then a3 divides b3.
Statement is True.
If a divides b
b/a = k, so b=ak
b3/a3 = (ak)3/a3 = a3k3/a3 = k3,
For each value of b and a its true.
(viii) If a2 ends in 9, then a3 ends in 7.
Statement is False.
Let a = 7
72 = 49 and 73 = 343
(ix) If a2 ends in an even number of zeros, then a3 ends in 25.
Statement is False.
Since, when a = 20
a2 = 202 = 400 and a3 = 8000 (a3 doesn’t end with 25)
(x) If a2 ends in an even number of zeros, then a3 ends in an odd number of zeros.
Statement is False.
Since, when a = 100
a2 = 1002 = 10000 and a3 = 1003 = 1000000 (a3 doesn’t end with odd number of zeros)
EXERCISE 4.2 PAGE NO: 4.13
1. Find the cubes of:
(i) 11
(ii) 12
(iii) 21
(i) 11
(ii) 12
(iii) 21
Solution:
(i) 11
The cube of 11 is
(11)3 = 11× 11× 11 = 1331
(ii) 12
The cube of 12 is
(12)3 = 12× 12× 12 = 1728
(iii) 21
The cube of 21 is
(21)3 = 21× 21× 21 = 9261
2. Which of the following integers are cubes of negative integers
(i) 64
(ii) 1056
(iii) 2197
(iv) 2744
(v) 42875
(i) 64
(ii) 1056
(iii) 2197
(iv) 2744
(v) 42875
Solution:
(i) 64
The prime factors of 64 are
64 = 2 × 2 × 2 × 2 × 2 × 2
= 23 × 23
= 43
∴ 64 is a perfect cube of negative integer – 4.
(ii) 1056
The prime factors of 1056 are
1056 = 2 × 2 × 2 × 2 × 2 × 3 × 11
1056 is not a perfect cube.
∴ 1056 is not a cube of negative integer.
(iii) 2197
The prime factors of 2197 are
2197 = 13 × 13 × 13
= 133
∴ 2197 is a perfect cube of negative integer – 13.
(iv) 2744
The prime factors of 2744 are
2744 = 2 × 2 × 2 × 7 × 7 × 7
= 23 × 73
= 143
2744 is a perfect cube.
∴ 2744 is a cube of negative integer – 14.
(v) 42875
The prime factors of 42875 are
42875 = 5 × 5 × 5 × 7 × 7 × 7
= 53 × 73
= 353
42875 is a perfect cube.
∴ 42875 is a cube of negative integer – 35.
3. Show that the following integers are cubes of negative integers. Also, find the integer whose cube is the given integer.
(i) 5832
(ii) 2744000
(i) 5832
(ii) 2744000
Solution:
(i) 5832
The prime factors of 5832 are
5823 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3
= 23 × 33 × 33
= 183
5832 is a perfect cube.
∴ 5832 is a cube of negative integer – 18.
(ii) 2744000
The prime factors of 2744000 are
2744000 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5 × 5 × 7 × 7 × 7
= 23 × 23× 53 × 73
2744000 is a perfect cube.
∴ 2744000 is a cube of negative integer – 140.
4. Find the cube of:
(i) 7/9 (ii) 8/11
(iii) 12/7 (iv) 13/8
(v) 2 2/5 (vi) 3 1/4
(vii) 0.3 (viii) 1.5
(ix) 0.08 (x) 2.1
(i) 7/9 (ii) 8/11
(iii) 12/7 (iv) 13/8
(v) 2 2/5 (vi) 3 1/4
(vii) 0.3 (viii) 1.5
(ix) 0.08 (x) 2.1
Solution:
(i) 7/9
The cube of 7/9 is
(7/9)3 = 73/93 = 343/729
(ii) 8/11
The cube of 8/11 is
(8/11)3 = 83/113 = 512/1331
(iii) 12/7
The cube of 12/7 is
(12/7)3 = 123/73 = 1728/343
(iv) 13/8
The cube of 13/8 is
(13/8)3 = 133/83 = 2197/512
(v) 2 2/5
The cube of 12/5 is
(12/5)3 = 123/53 = 1728/125
(vi) 3 ¼
The cube of 13/4 is
(13/4)3 = 133/43 = 2197/64
(vii) 0.3
The cube of 0.3 is
(0.3)3 = 0.3×0.3×0.3 = 0.027
(viii) 1.5
The cube of 1.5 is
(1.5)3 = 1.5×1.5×1.5 = 3.375
(ix) 0.08
The cube of 0.08 is
(0.08)3 = 0.08×0.08×0.08 = 0.000512
(x) 2.1
The cube of 2.1 is
(2.1)3 = 2.1×2.1×2.1 = 9.261
5. Find which of the following numbers are cubes of rational numbers:
(i) 27/64
(ii) 125/128
(iii) 0.001331
(iv) 0.04
(i) 27/64
(ii) 125/128
(iii) 0.001331
(iv) 0.04
Solution:
(i) 27/64
We have,
27/64 = (3×3×3)/ (4×4×4) = 33/43 = (3/4)3
∴ 27/64 is a cube of 3/4.
(ii) 125/128
We have,
125/128 = (5×5×5)/ (2×2×2×2×2×2×2) = 53/ (23×23×2)
∴ 125/128 is not a perfect cube.
(iii) 0.001331
We have,
1331/1000000 = (11×11×11)/ (100×100×100) = 113/1003 = (11/100)3
∴ 0.001331 is a perfect cube of 11/100
(iv) 0.04
We have,
4/10 = (2×2)/(2×5) = 22/(2×5)
∴ 0.04 is not a perfect cube.
EXERCISE 4.3 PAGE NO: 4.21
1. Find the cube roots of the following numbers by successive subtraction of numbers:
1, 7, 19, 37, 61, 91, 127, 169, 217, 271, 331, 397, …
(i) 64
(ii) 512
(iii) 1728
1, 7, 19, 37, 61, 91, 127, 169, 217, 271, 331, 397, …
(i) 64
(ii) 512
(iii) 1728
Solution:
(i) 64
Let’s perform subtraction
64 – 1 = 63
63 – 7 = 56
56 – 19 =37
37 – 37 = 0
Subtraction is performed 4 times.
∴ Cube root of 64 is 4.
(ii) 512
Let’s perform subtraction
512 – 1 = 511
511 – 7 = 504
504 – 19 = 485
485 – 37 = 448
448 – 61 = 387
387 – 91 = 296
296 – 127 = 169
169 – 169 = 0
Subtraction is performed 8 times.
∴ Cube root of 512 is 8.
(iii) 1728
Let’s perform subtraction
1728 – 1 = 1727
1727 – 7 = 1720
1720 – 19 = 1701
1701 – 37 = 1664
1664 – 91 = 1512
1512 – 127 = 1385
1385 – 169 = 1216
1216 – 217 = 999
999 – 271 = 728
728 – 331 = 397
397 – 397 = 0
Subtraction is performed 12 times.
∴ Cube root of 1728 is 12.
2. Using the method of successive subtraction examine whether or not the following numbers are perfect cubes:
(i) 130
(ii) 345
(iii) 792
(iv) 1331
(i) 130
(ii) 345
(iii) 792
(iv) 1331
Solution:
(i) 130
Let’s perform subtraction
130 – 1 = 129
129 – 7 = 122
122 – 19 = 103
103 – 37 = 66
66 – 61 = 5
Next number to be subtracted is 91, which is greater than 5
∴130 is not a perfect cube.
(ii) 345
Let’s perform subtraction
345 – 1 = 344
344 – 7 = 337
337 – 19 = 318
318 – 37 = 281
281 – 61 = 220
220 – 91 = 129
129 – 127 = 2
Next number to be subtracted is 169, which is greater than 2
∴ 345 is not a perfect cube
(iii) 792
Let’s perform subtraction
792 – 1 = 791
791 – 7 = 784
784 – 19 = 765
765 – 37 = 728
728 – 61 = 667
667 – 91 = 576
576 – 127 = 449
449 – 169 = 280
280 – 217 = 63
Next number to be subtracted is 271, which is greater than 63
∴ 792 is not a perfect cube
(iv) 1331
Let’s perform subtraction
1331 – 1 = 1330
1330 – 7 = 1323
1323 – 19 = 1304
1304 – 37 = 1267
1267 – 61 = 1206
1206 – 91 = 1115
1115 – 127 = 988
988 – 169 = 819
819 – 217 = 602
602 – 271 = 331
331 – 331 = 0
Subtraction is performed 11 times
Cube root of 1331 is 11
∴ 1331 is a perfect cube.
3. Find the smallest number that must be subtracted from those of the numbers in question 2 which are not perfect cubes, to make them perfect cubes. What are the corresponding cube roots?
Solution:
In previous question there are three numbers which are not perfect cubes.
(i) 130
Let’s perform subtraction
130 – 1 = 129
129 – 7 = 122
122 – 19 = 103
103 – 37 = 66
66 – 61 = 5
Next number to be subtracted is 91, which is greater than 5
Since, 130 is not a perfect cube. So, to make it perfect cube we subtract 5 from the given number.
130 – 5 = 125 (which is a perfect cube of 5)
(ii) 345
Let’s perform subtraction
345 – 1 = 344
344 – 7 = 337
337 – 19 = 318
318 – 37 = 281
281 – 61 = 220
220 – 91 = 129
129 – 127 = 2
Next number to be subtracted is 169, which is greater than 2
Since, 345 is not a perfect cube. So, to make it a perfect cube we subtract 2 from the given number.
345 – 2 = 343 (which is a perfect cube of 7)
(iii) 792
Let’s perform subtraction
792 – 1 = 791
791 – 7 = 784
784 – 19 = 765
765 – 37 = 728
728 – 61 = 667
667 – 91 = 576
576 – 127 = 449
449 – 169 = 280
280 – 217 = 63
Next number to be subtracted is 271, which is greater than 63
Since, 792 is not a perfect cube. So, to make it a perfect cube we subtract 63 from the given number.
792 – 63 = 729 (which is a perfect cube of 9)
4. Find the cube root of each of the following natural numbers:
(i) 343 (ii) 2744
(i) 343 (ii) 2744
(iii) 4913 (iv) 1728
(v) 35937 (vi) 17576
(vii) 134217728 (viii) 48228544
(ix) 74088000 (x) 157464
(xi) 1157625 (xii) 33698267
(v) 35937 (vi) 17576
(vii) 134217728 (viii) 48228544
(ix) 74088000 (x) 157464
(xi) 1157625 (xii) 33698267
Solution:
(i) 343
By using prime factorization method
∛343 = ∛ (7×7×7) = 7
(ii) 2744
By using prime factorization method
∛2744 = ∛ (2×2×2×7×7×7) = ∛ (23×73) = 2×7 = 14
(iii) 4913
By using prime factorization method,
∛4913 = ∛ (17×17×17) = 17
(iv) 1728
By using prime factorization method,
∛1728 = ∛(2×2×2×2×2×2×3×3×3) = ∛ (23×23×33) = 2×2×3 = 12
(v) 35937
By using prime factorization method,
∛35937 = ∛ (3×3×3×11×11×11) = ∛ (33×113) = 3×11 = 33
(vi) 17576
By using prime factorization method,
∛17576 = ∛ (2×2×2×13×13×13) = ∛ (23×133) = 2×13 = 26
(vii) 134217728
By using prime factorization method
∛134217728 = ∛ (227) = 29 = 512
(viii) 48228544
By using prime factorization method
∛48228544 = ∛ (2×2×2×2×2×2×7×7×7×13×13×13) = ∛ (23×23×73×133) = 2×2×7×13 = 364
(ix) 74088000
By using prime factorization method
∛74088000 = ∛ (2×2×2×2×2×2×3×3×3×5×5×5×7×7×7) = ∛ (23×23×33×53×73) = 2×2×3×5×7 = 420
(x) 157464
By using prime factorization method
∛157464 = ∛ (2×2×2×3×3×3×3×3×3×3×3×3) = ∛ (23×33×33×33) = 2×3×3×3 = 54
(xi) 1157625
By using prime factorization method
∛1157625 = ∛ (3×3×3×5×5×5×7×7×7) = ∛ (33×53×73) = 3×5×7 = 105
(xii) 33698267
By using prime factorization method
∛33698267 = ∛ (17×17×17×19×19×19) = ∛ (173×193) = 17×19 = 323
5. Find the smallest number which when multiplied with 3600 will make the product a perfect cube. Further, find the cube root of the product.
Solution:
Firstly let’s find the prime factors for 3600
3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5
= 23 × 32 × 52 × 2
Since only one triples is formed and three factors remained ungrouped in triples.
The given number 3600 is not a perfect cube.
To make it a perfect cube we have to multiply it by (2 × 2 × 3 × 5) = 60
3600 × 60 = 216000
Cube root of 216000 is
∛216000 = ∛ (60×60×60) = ∛ (603) = 60
∴ the smallest number which when multiplied with 3600 will make the product a perfect cube is 60 and the cube root of the product is 60.
6. Multiply 210125 by the smallest number so that the product is a perfect cube. Also, find out the cube root of the product.
Solution:
The prime factors of 210125 are
210125 = 5 × 5 × 5 × 41 × 41
Since, one triples remained incomplete, 210125 is not a perfect cube.
To make it a perfect cube we need to multiply the factors by 41, we will get 2 triples as 23 and 413.
And the product become:
210125 × 41 = 8615125
8615125 = 5 × 5 × 5 × 41 × 41 × 41
Cube root of product = ∛8615125 = ∛ (5×41) = 205
7. What is the smallest number by which 8192 must be divided so that quotient is a perfect cube? Also, find the cube root of the quotient so obtained.
Solution:
The prime factors of 8192 are
8192 = 2×2×2×2×2×2×2×2×2×2×2 = 23×23×23×2
Since, one triples remain incomplete, hence 8192 is not a perfect cube.
So, we divide 8192 by 2 to make its quotient a perfect cube.
8192/2 = 4096
4096 = 2×2×2×2×2×2×2×2×2×2×2×2 = 23×23×23×23
Cube root of 4096 = ∛4096 = ∛ (23×23×23×23) = 2×2×2×2 = 16
8. Three numbers are in the ratio 1:2:3. The sum of their cubes is 98784. Find the numbers.
Solution:
Let us consider the ratio 1:2:3 as x, 2x and 3x
According to the question,
X3 + (2x) 3 + (3x) 3 = 98784
x3 + 8x3 + 27x3 = 98784
36x3 = 98784
x3 = 98784/36
= 2744
x = ∛2744 = ∛ (2×2×2×7×7×7) = 2×7 = 14
So, the numbers are,
x = 14
2x = 2 × 14 = 28
3x = 3 × 14 = 42
9. The volume of a cube is 9261000 m3. Find the side of the cube.
Given, volume of cube = 9261000 m3
Let us consider the side of cube be ‘a’ metre
So, a3 = 9261000
a = ∛9261000 = ∛ (2×2×2×3×3×3×5×5×5×7×7×7) = ∛ (23×33×53×73) = 2×3×5×7 = 210
∴ the side of cube = 210 metre
EXERCISE 4.4 PAGE NO: 4.30
1. Find the cube roots of each of the following integers:
(i)125 (ii) 5832
(iii)2744000 (iv) 753571
(v) 32768
(i)125 (ii) 5832
(iii)2744000 (iv) 753571
(v) 32768
Solution:
(i) 125
The cube root of 125 is
125 = ∛125 = ∛125 = ∛ (5×5×5) = 5
(ii) 5832
The cube root of 5832 is
5832 = ∛5832 = ∛5832
To find the cube root of 5832, we shall use the method of unit digits.
Let us consider the number 5832. Where, unit digit of 5832 = 2
Unit digit in the cube root of 5832 will be 8
After striking out the units, tens and hundreds digits of 5832,
Now we left with 5 only.
We know that 1 is the Largest number whose cube is less than or equal to 5.
So, the tens digit of the cube root of 5832 is 1.
∛5832 = ∛5832 = 18
(iii) 2744000
∛2744000 = ∛2744000
We shall use the method of factorization to find the cube root of 2744000
So let’s find the prime factors for 2744000
2744000 = 2×2×2×2×2×2×5×5×5×7×7×7
Now by grouping the factors into triples of equal factors, we get,
2744000 = (2×2×2) × (2×2×2) × (5×5×5) × (7×7×7)
Since all the prime factors of 2744000 is grouped in to triples of equal factors and no factor is left over.
So now take one factor from each group and by multiplying we get,
2×2×5×7 = 140
Thereby we can say that 2744000 is a cube of 140
∴ ∛2744000 = ∛2744000 = 140
(iv) 753571
∛753571 = ∛753571
We shall use the unit digit method,
Let us consider the number 753571, where unit digit = 1
Unit digit in the cube root of 753571 will be 1
After striking out the units, tens and hundreds digits of 753571,
Now we left with 753.
We know that 9 is the Largest number whose cube is less than or equal to 753(93<753 span="" style="boxsizing: borderbox; fontsize: 10.5px; lineheight: 0; position: relative; top: 0.5em; verticalalign: baseline;">3
So, the tens digit of the cube root of 753571 is 9.
∛753571 = 91
∛753571 = ∛753571 = 91
(v) 32768
∛32765 = ∛32768
We shall use the unit digit method,
Let us consider the Number = 32768, where unit digit = 8
Unit digit in the cube root of 32768 will be 2
After striking out the units, tens and hundreds digits of 32768,
Now we left with 32.
As we know that 9 is the Largest number whose cube is less than or equals to 32(33<32 span="" style="boxsizing: borderbox; fontsize: 10.5px; lineheight: 0; position: relative; top: 0.5em; verticalalign: baseline;">3
).
So, the tens digit of the cube root of 32768 is 3.
∛32765 = 32
∛32765 = ∛32768 = 32
2. Show that:
(i) ∛27 × ∛64 = ∛ (27×64)
(ii) ∛ (64×729) = ∛64 × ∛729
(iii) ∛ (125×216) = ∛125 × ∛216
(iv) ∛ (125×1000) = ∛125 × ∛1000
Solution:
(i) ∛27 × ∛64 = ∛ (27×64)
Let us consider LHS ∛27 × ∛64
∛27 × ∛64 = ∛(3×3×3) × ∛(4×4×4)
= 3×4
= 12
Let us consider RHS ∛ (27×64)
∛ (27×64) = ∛ (3×3×3×4×4×4)
= 3×4
= 12
∴ LHS = RHS, the given equation is verified.
(ii) ∛ (64×729) = ∛64 × ∛729
Let us consider LHS ∛ (64×729)
∛ (64×729) = ∛ (4×4×4×9×9×9)
= 4×9
= 36
Let us consider RHS ∛64 × ∛729
∛64 × ∛729 = ∛(4×4×4) × ∛(9×9×9)
= 4×9
= 36
∴ LHS = RHS, the given equation is verified.
(iii) ∛ (125×216) = ∛125 × ∛216
Let us consider LHS ∛ (125×216)
∛ (125×216) = ∛ (5×5×5×2×2×2×3×3×3)
= 5×2×3
= 30
Let us consider RHS ∛125 × ∛216
∛125 × ∛216 = ∛(5×5×5) × ∛(2×2×2×3×3×3)
= 5×2×3
= 30
∴ LHS = RHS, the given equation is verified.
(iv) ∛ (125×1000) = ∛125 × ∛1000
Let us consider LHS ∛ (125×1000)
∛ (125×1000) = ∛ (5×5×5×10×10×10)
= 5×10
= 50
Let us consider RHS ∛125 × ∛1000
∛125 × ∛1000 = ∛(5×5×5) × ∛(10×10×10)
= 5×10
= 50
∴ LHS = RHS, the given equation is verified.
3. Find the cube root of each of the following numbers:
(i) 8×125
(i) 8×125
(ii) 1728×216
(iii) 27×2744
(iii) 27×2744
(iv) 729×15625
Solution:
(i) 8×125
We know that for any two integers a and b, ∛ (a×b) = ∛a × ∛b
By using the property
∛ (8×125) = ∛8 × ∛125
= ∛ (2×2×2) × ∛ (5×5×5)
= 2×5
= 10
(ii) 1728×216
We know that for any two integers a and b, ∛ (a×b) = ∛a × ∛b
By using the property
∛ (1728×216) = ∛1728 × ∛216
We shall use the unit digit method
Let us consider the number 1728, where Unit digit = 8
The unit digit in the cube root of 1728 will be 2
After striking out the units, tens and hundreds digits of the given number, we are left with the 1.
We know 1 is the largest number whose cube is less than or equal to 1.
So, the tens digit of the cube root of 1728 = 1
∛1728 = 12
Now, let’s find the prime factors for 216 = 2×2×2×3×3×3
By grouping the factors in triples of equal factor, we get,
216 = (2×2×2) × (3×3×3)
By taking one factor from each group we get,
∛216 = 2×3 = 6
∴ by equating the values in the given equation we get,
∛ (1728×216) = ∛1728 × ∛216
= 12 × 6
= 72
(iii) 27×2744
We know that for any two integers a and b, ∛ (a×b) = ∛a × ∛b
By using the property
∛ (27×2744) = ∛27 × ∛2744
We shall use the unit digit method
Let us consider the number 2744, where Unit digit = 4
The unit digit in the cube root of 2744 will be 4
After striking out the units, tens and hundreds digits of the given number, we are left with the 2.
We know 2 is the largest number whose cube is less than or equal to 2.
So, the tens digit of the cube root of 2744 = 2
∛2744 = 14
Now, let’s find the prime factors for 27 = 3×3×3
By grouping the factors in triples of equal factor, we get,
27 = (3×3×3)
Cube root of 27 is
∛27 = 3
∴ by equating the values in the given equation we get,
∛ (27×2744) = ∛27 × ∛2744
= 3 × 14
= 42
(iv) 729×15625
We know that for any two integers a and b, ∛ (a×b) = ∛a × ∛b
By using the property
∛ (729×15625) = ∛729 × ∛15625
We shall use the unit digit method
Let us consider the number 15625, where Unit digit = 5
The unit digit in the cube root of 15625 will be 5
After striking out the units, tens and hundreds digits of the given number, we are left with the 15.
We know 15 is the largest number whose cube is less than or equal to 15(23<15 span="" style="boxsizing: borderbox; fontsize: 10.5px; lineheight: 0; position: relative; top: 0.5em; verticalalign: baseline;">3
).
So, the tens digit of the cube root of 15625 = 2
∛15625 = 25
Now, let’s find the prime factors for 729 = 9×9×9
By grouping the factors in triples of equal factor, we get,
729 = (9×9×9)
Cube root of 729 is
∛729 = 9
∴ by equating the values in the given equation we get,
∛ (729×15625) = ∛729 × ∛15625
= 9 × 25
= 225
4. Evaluate:
(i) ∛ (43 × 63)
(ii) ∛ (8×17×17×17)
(iii) ∛ (700×2×49×5)
(iv) 125 ∛a6 – ∛125a6
Solution:
(i) ∛ (43 × 63)
We know that for any two integers a and b, ∛ (a×b) = ∛a × ∛b
By using the property
∛ (43 × 63) = ∛43 × ∛63
= 4 × 6
= 24
(ii) ∛ (8×17×17×17)
We know that for any two integers a and b, ∛ (a×b) = ∛a × ∛b
By using the property
∛ (8×17×17×17) = ∛8 × ∛17×17×17
= ∛23 × ∛173
= 2 × 17
= 34
(iii) ∛ (700×2×49×5)
Firstly let us find the prime factors for the above numbers
∛ (700×2×49×5) = ∛ (2×2×5×5×7×2×7×7×5)
= ∛ (23×53×73)
= 2×5×7
= 70
(iv) 125 ∛a6 – ∛125a6
125 ∛a6 – ∛125a6 = 125 ∛(a2)3 – ∛53(a2)3
= 125a2 – 5a2
= 120a2
5. Find the cube root of each of the following rational numbers:
(i) 125/729
(ii) 10648/12167
(iii) 19683/24389
(iv) 686/3456
(v) 39304/42875
Solution:
(i) 125/729
Let us find the prime factors of 125 and 729
125/729 = – (∛ (5×5×5)) / (∛ (9×9×9))
= – (∛ (53)) / (∛ (93))
= – 5/9
(ii) 10648/12167
Let us find the prime factors of 10648 and 12167
10648/12167 = (∛ (2×2×2×11×11×11)) / (∛ (23×23×23))
= (∛ (23×113)) / (∛ (233))
= (2×11)/23
= 22/23
(iii) 19683/24389
Let us find the prime factors of 19683 and 24389
19683/24389 = (∛ (3×3×3×3×3×3×3×3×3)) / (∛ (29×29×29))
= – (∛ (33×33×33)) / (∛ (293))
= – (3×3×3)/29
= – 27/29
(iv) 686/3456
Let us find the prime factors of 686 and 3456
686/3456 = = (∛ (2×7×7×7)) / (∛ (27×23))
= – (∛ (2×73)) / (∛ (27×23))
= – (∛ (73)) / (∛ (26×23))
= – 7/(2×2×2)
= – 7/8
(v) 39304/42875
Let us find the prime factors of 39304 and 42875
39304/42875 = (∛ (2×2×2×17×17×17)) / (∛ (5×5×5×7×7×7))
= – (∛ (23×173)) / (∛ (53×73))
= – (2×17)/( 5×7)
= – 34/35
= 34/35
6. Find the cube root of each of the following rational numbers:
(i) 0.001728
(ii) 0.003375
(iii) 0.001
(iv) 1.331
(i) 0.001728
(ii) 0.003375
(iii) 0.001
(iv) 1.331
Solution:
(i) 0.001728
0.001728 = 1728/1000000
∛ (0.001728) = ∛1728 / ∛1000000
Let us find the prime factors of 1728 and 1000000
∛(0.001728) = ∛(23×23×33) / ∛(1003)
= (2×2×3)/100
= 12/100
= 0.12
(ii) 0.003375
0.003375 = 3375/1000000
∛ (0.003375) = ∛3375 / ∛1000000
Let us find the prime factors of 3375 and 1000000
∛(0.003375) = ∛(33×53) / ∛(1003)
= (3×5)/100
= 15/100
= 0.15
(iii) 0.001
0.001 = 1/1000
∛ (0.001) = ∛1 / ∛1000
= 1/ ∛103
= 1/10
= 0.1
(iv) 1.331
1.331 = 1331/1000
∛ (1.331) = ∛1331 / ∛1000
Let us find the prime factors of 1331 and 1000
∛(1.331) = ∛(113) / ∛(103)
= 11/10
= 1.1
7. Evaluate each of the following:
(i) ∛27 + ∛0.008 + ∛0.064
(ii) ∛1000 + ∛0.008 – ∛0.125
(iii) ∛(729/216) × 6/9
(iv) ∛(0.027/0.008) ÷ ∛(0.09/0.04) – 1
(v) ∛(0.1×0.1×0.1×13×13×13)
(v) ∛(0.1×0.1×0.1×13×13×13)
Solution:
(i) ∛27 + ∛0.008 + ∛0.064
Let us simplify
∛ (3×3×3) + ∛ (0.2×0.2×0.2) + ∛ (0.4×0.4×0.4)
∛ (3)3 + ∛ (0.2)3 + ∛ (0.4)3
3 + 0.2 + 0.4
3.6
(ii) ∛1000 + ∛0.008 – ∛0.125
Let us simplify
∛ (10×10×10) + ∛ (0.2×0.2×0.2) – ∛ (0.5×0.5×0.5)
∛ (10)3 + ∛ (0.2)3 – ∛ (0.5)3
10 + 0.2 – 0.5
9.7
(iii) ∛ (729/216) × 6/9
Let us simplify
∛ (9×9×9/6×6×6) × 6/9
(∛ (9)3 / ∛ (6)3)× 6/9
9/6 × 6/9
1
(iv) ∛(0.027/0.008) ÷ √(0.09/0.04) – 1
Let us simplify ∛(0.027/0.008) ÷ √ (0.09/0.04)
∛(0.3×0.3×0.3/0.2×0.2×0.2) ÷ √(0.3×0.3/0.2×0.2)
(∛(0.3)3 / ∛(0.2)3) ÷ (√(0.3)2 / √(0.2)2)
(0.3/0.2) ÷ (0.3/0.2) 1
(0.3/0.2 × 0.2/0.3) 1
1 – 1
0
(v) ∛(0.1×0.1×0.1×13×13×13)
∛(0.13×133)
0.1 × 13 = 1.3
8. Show that:
(i) ∛ (729)/ ∛ (1000) = ∛ (729/1000)
(ii) ∛ (512)/ ∛ (343) = ∛ (512/343)
Solution:
(i) ∛ (729)/ ∛ (1000) = ∛ (729/1000)
Let us consider LHS ∛ (729)/ ∛ (1000)
∛ (729)/ ∛ (1000) = ∛ (9×9×9)/ ∛ (10×10×10)
= ∛ (93/103)
= 9/10
Let us consider RHS ∛ (729/1000)
∛ (729/1000) = ∛ (9×9×9/10×10×10)
= ∛ (93/103)
= 9/10
∴ LHS = RHS
(ii) ∛ (512)/ ∛ (343) = ∛ (512/343)
Let us consider LHS ∛ (512)/ ∛ (343)
∛ (512)/ ∛ (343) = ∛(8×8×8)/ ∛ (7×7×7)
= ∛(83/73)
= 8/7
Let us consider RHS ∛ (512/343)
∛ (512/343) = = ∛(8×8×8/7×7×7)
= ∛(83/73)
= 8/7
∴ LHS = RHS
9. Fill in the blanks:
(i) ∛(125×27) = 3 × …
(ii) ∛(8×…) = 8
(iii) ∛1728 = 4 × …
(iii) ∛1728 = 4 × …
(iv) ∛480 = ∛3×2× ∛..
(v) ∛… = ∛7 × ∛8
(vi) ∛..= ∛4 × ∛5 × ∛6
(vii) ∛(27/125) = …/5
(viii) ∛(729/1331) = 9/…
(ix) ∛(512/…) = 8/13
Solution:
(i) ∛(125×27) = 3 × …
Let us consider LHS ∛(125×27)
∛(125×27) = ∛(5×5×5×3×3×3)
= ∛(53×33)
= 5×3 or 3×5
(ii) ∛(8×…) = 8
Let us consider LHS ∛(8×…)
∛(8×8×8) = ∛83 = 8
(iii) ∛1728 = 4 × …
Let us consider LHS
∛1728 = ∛(2×2×2×2×2×2×3×3×3)
= ∛(23×23×33)
= 2×2×3
= 4×3
(iv) ∛480 = ∛3×2× ∛..
Let us consider LHS
∛480 = ∛(2×2×2×2×2×3×5)
= ∛(23×22×3×5)
= ∛23× ∛3 × ∛2×2×5
= 2 × ∛3 × ∛20
(v) ∛… = ∛7 × ∛8
Let us consider RHS
∛7 × ∛8 = ∛(7 × 8)
= ∛56
(vi) ∛..= ∛4 × ∛5 × ∛6
Let us consider RHS
∛4 × ∛5 × ∛6 = ∛(4 × 5 × 6)
= ∛120
(vii) ∛(27/125) = …/5
Let us consider LHS
∛(27/125) = ∛(3×3×3)/(5×5×5)
= ∛(33)/(53)
= 3/5
(viii) ∛(729/1331) = 9/…
Let us consider LHS
∛(729/1331) = ∛(9×9×9)/(11×11×11)
= ∛(93)/(113)
= 9/11
(ix) ∛(512/…) = 8/13
Let us consider LHS
∛(512/…) = ∛(2×2×2×2×2×2×2×2×2)
= ∛(23×23×23)
= 2×2×2
= 8
So, 8/∛… = 8/13
when numerators are same the denominators will also become equal.
8 × 13 = 8 × ∛…
∛…= 13
… = (13)3
= 2197
10. The volume of a cubical box is 474. 552 cubic metres. Find the length of each side of the box.
Solution:
Volume of a cubical box is 474.552 cubic metres
V = 83,
Let ‘S’ be the side of the cube
83 = 474.552 cubic metres
8 = ∛474.552
= ∛ (474552/1000)
Let us factorise 474552 into prime factors, we get:
474552 = 2×2×2×3×3×3×13×13×13
By grouping the factors in triples of equal factors, we get:
474552 = (2×2×2) × (3×3×3) × (13×13×13)
Now, ∛474.552 = ∛ ((2×2×2) × (3×3×3) × (13×13×13))
= 2×3×13
= 78
Also,
∛1000 = ∛ (10×10×10)
= ∛ (10)3
= 10
So now let us equate in the above equation we get,
8 = ∛ (474552/1000)
= 78/10
= 7.8
∴ length of the side is 7.8m.
11. Three numbers are to one another 2:3:4. The sum of their cubes is 0.334125. Find the numbers.
Solution:
Let us consider the ratio 2:3:4 be 2a, 3a, and 4a.
So according to the question:
(2a)3+ (3a)3+(4a)3 = 0.334125
8a3+27 a3+64 a3 = 0.334125
99a3 = 0.334125
a3 = 334125/1000000×99
= 3375/1000000
a = ∛ (3375/1000000)
= ∛((15×15×15)/ 100×100×100)
= 15/100
= 0.15
∴ The numbers are:
2×0.15 = 0.30
3×0.15 = 0.45
4×0.15 = 0.6
12. Find the side of a cube whose volume is 24389/216m3.
Solution:
Volume of the side s = 24389/216 = v
V= 8.3
8 = ∛v
= ∛(24389/216)
By performing factorisation we get,
= ∛(29×29×29/2×2×2×3×3×3)
= 29/(2×3)
= 29/6
∴ The length of the side is 29/6.
13. Evaluate:
(i) ∛36 × ∛384
(ii) ∛96 × ∛144
(iii) ∛100 × ∛270
(iv) ∛121 × ∛297
Solution:
(i) ∛36 × ∛384
We know that ∛a × ∛b = ∛ (a×b)
By using the above formula let us simplify
∛36 × ∛384 = ∛ (36×384)
The prime factors of 36 and 384 are
= ∛ (2×2×3×3) × (2×2×2×2×2×2×3×3×3)
= ∛(23×23×23×33)
= 2×2×2×3
= 24
(ii) ∛96 × ∛144
We know that ∛a × ∛b = ∛ (a×b)
By using the above formula let us simplify
∛96 × ∛144 = ∛ (96×144)
The prime factors of 96 and 144 are
= ∛ (2×2×2×2×2×3) × (2×2×2×2×3×3)
= ∛(23×23×23×33)
= 2×2×2×3
= 24
(iii) ∛100 × ∛270
We know that ∛a × ∛b = ∛ (a×b)
By using the above formula let us simplify
∛100 × ∛270 = ∛ (100×270)
The prime factors of 100 and 270 are
= ∛ (2×2×5×5) × (2×3×3×3×5)
= ∛(23×33×53)
= 2×3×5
= 30
(iv) ∛121 × ∛297
We know that ∛a × ∛b = ∛ (a×b)
By using the above formula let us simplify
∛121 × ∛297 = ∛ (121×297)
The prime factors of 121 and 297 are
= ∛ (11×11) × (3×3×3×11)
= ∛(113×33)
= 11×3
= 33
14. Find the cube roots of the numbers 3048625, 20346417, 210644875, 57066625 using the fact that
(i) 3048625 = 3375 × 729
(ii) 20346417 = 9261 × 2197
(iii) 210644875 = 42875 × 4913
(iv) 57066625 = 166375 × 343
(i) 3048625 = 3375 × 729
(ii) 20346417 = 9261 × 2197
(iii) 210644875 = 42875 × 4913
(iv) 57066625 = 166375 × 343
Solution:
(i) 3048625 = 3375 × 729
By taking the cube root for the whole we get,
∛3048625 = ∛3375 × ∛729
Now perform factorization
= ∛3×3×3×5×5×5 × ∛9×9×9
= ∛33×53 × ∛93
= 3×5×9
= 135
(ii) 20346417 = 9261 × 2197
By taking the cube root for the whole we get,
∛20346417 = ∛9261 × ∛2197
Now perform factorization
= ∛3×3×3×7×7×7 × ∛13×13×13
= ∛33×73 × ∛133
= 3×7×13
= 273
(iii) 210644875 = 42875 × 4913
By taking the cube root for the whole we get,
∛210644875 = ∛42875 × ∛4913
Now perform factorization
= ∛5×5×5×7×7×7 × ∛17×17×17
= ∛53×73 × ∛173
= 5×7×17
= 595
(iv) 57066625 = 166375 × 343
By taking the cube root for the whole we get,
∛57066625 = ∛166375 × ∛343
Now perform factorization
= ∛5×5×5×11×11×11 × ∛7×7×7
= ∛53×113 × ∛73
= 5×11×7
= 385
15. Find the unit of the cube
root of the following numbers:
(i) 226981
(ii) 13824
(iii) 571787
(iv) 175616
root of the following numbers:
(i) 226981
(ii) 13824
(iii) 571787
(iv) 175616
Solution:
(i) 226981
The given number is 226981.
Unit digit of 226981 = 1
The unit digit of the cube root of 226981 = 1
(ii) 13824
The given number is 13824.
Unit digit of 13824 = 4
The unit digit of the cube root of 13824 = 4
(iii) 571787
The given number is 571787.
Unit digit of 571787 = 7
The unit digit of the cube root of 571787 = 7
(iv) 175616
The given number is 175616.
Unit digit of 175616 = 6
The unit digit of the cube root of 175616 = 6
16. Find the tens digit of the cube root of each of the numbers in Q.No.15.
(i) 226981
(ii) 13824
(iii) 571787
(iv) 175616
(i) 226981
(ii) 13824
(iii) 571787
(iv) 175616
Solution:
(i) 226981
The given number is 226981.
Unit digit of 226981 = 1
The unit digit in the cube root of 226981 = 1
After striking out the units, tens and hundreds digits of 226981, now we left with 226 only.
We know that 6 is the Largest number whose cube root is less than or equal to 226(63<226 span="" style="boxsizing: borderbox; fontsize: 10.5px; lineheight: 0; position: relative; top: 0.5em; verticalalign: baseline;">3
).
∴ The tens digit of the cube root of 226981 is 6.
(ii) 13824
The given number is 13824.
Unit digit of 13824 = 4
The unit digit in the cube root of 13824 = 4
After striking out the units, tens and hundreds digits of 13824, now we left with 13 only.
We know that 2 is the Largest number whose cube root is less than or equal to 13(23<13 span="" style="boxsizing: borderbox; fontsize: 10.5px; lineheight: 0; position: relative; top: 0.5em; verticalalign: baseline;">3
).
∴ The tens digit of the cube root of 13824 is 2.
(iii) 571787
The given number is 571787.
Unit digit of 571787 = 7
The unit digit in the cube root of 571787 = 3
After striking out the units, tens and hundreds digits of 571787, now we left with 571 only.
We know that 8 is the Largest number whose cube root is less than or equals to 571(83<571 span="" style="boxsizing: borderbox; fontsize: 10.5px; lineheight: 0; position: relative; top: 0.5em; verticalalign: baseline;">3
).
∴ The tens digit of the cube root of 571787 is 8.
(iv) 175616
The given number is 175616.
Unit digit of 175616 = 6
The unit digit in the cube root of 175616 = 6
After striking out the units, tens and hundreds digits of 175616, now we left with 175 only.
We know that 5 is the Largest number whose cube root is less than or equals to 175(53<175 span="" style="boxsizing: borderbox; fontsize: 10.5px; lineheight: 0; position: relative; top: 0.5em; verticalalign: baseline;">3
).
∴ The tens digit of the cube root of 175616 is 5.
EXERCISE 4.5 PAGE NO: 4.36
Making use of the cube root table, find the cube root of the following (correct to three decimal places):
1. 7
Solution:
As we know that 7 lies between 1 and 100 so by using cube root table we get,
∛7 = 1.913
∴ the answer is 1.913
2. 70
Solution:
As we know that 70 lies between 1 and 100 so by using cube root table from column x
We get,
∛70 = 4.121
∴ the answer is 4.121
3. 700
Solution:
700 = 70×10
By using cube root table 700 will be in the column ∛10x against 70.
So we get,
∛700 = 8.879
∴ the answer is 8.879
4. 7000
Solution:
7000 = 70×100
∛7000 = ∛(7×1000) = ∛7 × ∛1000
By using cube root table,
We get,
∛7 = 1.913
∛1000 = 10
∛7000 = ∛7 × ∛1000
= 1.913 × 10
= 19.13
∴ the answer is 19.13
5. 1100
Solution:
1100 = 11×100
∛1100 = ∛(11×100) = ∛11 × ∛100
By using cube root table,
We get,
∛11 = 2.224
∛100 = 4.6642
∛1100 = ∛11 × ∛100
= 2.224 × 4.642
= 10.323
∴ the answer is 10.323
6.780
Solution:
780 = 78×10
By using cube root table 780 would be in column ∛10x against 78.
We get,
∛780 = 9.205
7. 7800
Solution:
7800 = 78×100
∛7800 = ∛(78×100) = ∛78 × ∛100
By using cube root table,
We get,
∛78 = 4.273
∛100 = 4.6642
∛7800 = ∛78 × ∛100
= 4.273 × 4.642
= 19.835
∴ the answer is 19.835
8. 1346
Solution:
Let us find the factors by using factorisation method,
We get,
1346 = 2×673
∛1346 = ∛(2×676) = ∛2 × ∛673
Since, 670<673 p="">
By using cube root table,
∛670 = 8.750
∛680 = 8.794
For the difference (680670) which is 10.
So the difference in the values = 8.794 – 8.750 = 0.044
For the difference (673670) which is 3.
So the difference in the values = (0.044/10) × 3 = 0.0132
∛673 = 8.750 + 0.013 = 8.763
∛1346 = ∛2 × ∛673
= 1.260 × 8.763
= 11.041
∴ the answer is 11.041
9. 250
Solution:
250 = 25×100
By using cube root table 250 would be in column ∛10x against 25.
We get,
∛250 = 6.3
∴ the answer is 6.3
10. 5112
Solution:
Let us find the factors by using factorisation method,
∛5112 = ∛2×2×2×3×3×71
= ∛23×32×71
= 2 × ∛32 × ∛71
= 2 × ∛9 × ∛71
From cube root table we get,
∛9 = 2.080
∛71 = 4.141
∛5112 = 2 × ∛9 × ∛71
= 2 × 2.080 × 4.141
= 17.227
∴ the answer is 17.227
11. 9800
Solution:
∛9800 = ∛98 × ∛100
From cube root table we get,
∛98 = 4.610
∛100 = 4.642
∛9800 = ∛98 × ∛100
= 4.610 × 4.642
= 21.40
∴ the answer is 21.40
12. 732
Solution:
∛732
We know that value of ∛732 will lie between ∛730 and ∛740
From cube root table we get,
∛730 = 9.004
∛740 = 9.045
By using unitary method,
Difference between the values (740 – 730 = 10)
So, the difference in cube root values will be = 9.045 – 9.004 = 0.041
Difference between the values (732 – 730 = 2)
So, the difference in cube root values will be = (0.041/10) ×2 = 0.008
∛732 = 9.004+0.008 = 9.012
∴ the answer is 9.012
13. 7342
Solution:
∛7342
We know that value of ∛7342 will lie between ∛7300 and ∛7400
From cube root table we get,
∛7300 = 19.39
∛7400 = 19.48
By using unitary method,
Difference between the values (7400 – 7300 = 100)
So, the difference in cube root values will be = 19.48 – 19.39 = 0.09
Difference between the values (7342 – 7300 = 42)
So, the difference in cube root values will be = (0.09/100) × 42 = 0.037
∛7342 = 19.39+0.037 = 19.427
∴ the answer is 19.427
14. 133100
Solution:
∛133100 = ∛ (1331×100)
= ∛1331 × ∛100
= ∛ 113 × ∛100
= 11 × ∛100
From cube root table we get,
∛100 = 4.462
∛133100 = 11 × ∛100
= 11 × 4.462
= 51.062
∴ the answer is 51.062
15. 37800
Solution:
∛37800
Firstly let us find the factors for 37800
∛37800 = ∛(2×2×2×3×3×3×175)
= ∛(23×33×175)
= 6 × ∛175
We know that value of ∛175 will lie between ∛170 and ∛180
From cube root table we get,
∛170 = 5.540
∛180 = 5.646
By using unitary method,
Difference between the values (180 – 170 = 10)
So, the difference in cube root values will be = 5.646 – 5.540 = 0.106
Difference between the values (175 – 170 = 5)
So, the difference in cube root values will be = (0.106/10) × 5 = 0.053
∛175 = 5.540 + 0.053 = 5.593
∛37800 = 6 × ∛175
= 6 × 5.593
= 33.558
∴ the answer is 33.558
16. 0.27
Solution:
∛0.27 = ∛(27/100) = ∛27/∛100
From cube root table we get,
∛27 = 3
∛100 = 4.642
∛0.27 = ∛27/∛100
= 3/4.642
= 0.646
∴ the answer is 0.646
17. 8.6
Solution:
∛8.6 = ∛(86/10) = ∛86/∛10
From cube root table we get,
∛86 = 4.414
∛10 = 2.154
∛8.6 = ∛86/∛10
= 4.414/2.154
= 2.049
∴ the answer is 2.049
18. 0.86
Solution:
∛0.86 = ∛(86/100) = ∛86/∛100
From cube root table we get,
∛86 = 4.414
∛100 = 4.642
∛8.6 = ∛86/∛100
= 4.414/4.642
= 0.9508
∴ the answer is 0.951
19. 8.65
Solution:
∛8.65 = ∛(865/100) = ∛865/∛100
We know that value of ∛865 will lie between ∛860 and ∛870
From cube root table we get,
∛860 = 9.510
∛870 = 9.546
∛100 = 4.642
By using unitary method,
Difference between the values (870 – 860 = 10)
So, the difference in cube root values will be = 9.546 – 9.510 = 0.036
Difference between the values (865 – 860 = 5)
So, the difference in cube root values will be = (0.036/10) × 5 = 0.018
∛865 = 9.510 + 0.018 = 9.528
∛8.65 = ∛865/∛100
= 9.528/4.642
= 2.0525
∴ the answer is 2.053
20. 7532
Solution:
∛7532
We know that value of ∛7532 will lie between ∛7500 and ∛7600
From cube root table we get,
∛7500 = 19.57
∛7600 = 19.66
By using unitary method,
Difference between the values (7600 – 7500 = 100)
So, the difference in cube root values will be = 19.66 – 19.57 = 0.09
Difference between the values (7532 – 7500 = 32)
So, the difference in cube root values will be = (0.09/100) × 32 = 0.029
∛7532 = 19.57 + 0.029 = 19.599
∴ the answer is 19.599
21. 833
Solution:
∛833
We know that value of ∛833 will lie between ∛830 and ∛840
From cube root table we get,
∛830 = 9.398
∛840 = 9.435
By using unitary method,
Difference between the values (840 – 830 = 10)
So, the difference in cube root values will be = 9.435 – 9.398 = 0.037
Difference between the values (833 – 830 = 3)
So, the difference in cube root values will be = (0.037/10) ×3 = 0.011
∛833 = 9.398+0.011 = 9.409
∴ the answer is 9.409
22. 34.2
Solution:
∛34.2 = ∛(342/10) = ∛342/∛10
We know that value of ∛342 will lie between ∛340 and ∛350
From cube root table we get,
∛340 = 6.980
∛350 = 7.047
∛10 = 2.154
By using unitary method,
Difference between the values (350 – 340 = 10)
So, the difference in cube root values will be = 7.047 – 6.980 = 0.067
Difference between the values (342 – 340 = 2)
So, the difference in cube root values will be = (0.067/10) × 2 = 0.013
∛342 = 6.980 + 0.013 = 6.993
∛34.2 = ∛342/∛10
= 6.993/2.154
= 3.246
∴ the answer is 3.246
23. What is the length of the side of a cube whose volume is 275 cm3. Make use of the table for the cube root.
Solution:
The given volume of the cube = 275cm3
Let us consider the side of the cube as ‘a’cm
a3 = 275
a = ∛275
We know that value of ∛275 will lie between ∛270 and ∛280
From cube root table we get,
∛270 = 6.463
∛280 = 6.542
By using unitary method,
Difference between the values (280 – 270 = 10)
So, the difference in cube root values will be = 6.542 – 6.463 = 0.079
Difference between the values (275 – 270 = 5)
So, the difference in cube root values will be = (0.079/10) × 5 = 0.0395
∛275 = 6.463 + 0.0395 = 6.5025
∴ the answer is 6.503cm