7. Cubes and Cube Roots – Class 8 Mathematics NCERT Chapter 7 Complete Guide (CBSE)
NCERT Class 8 Mathematics Chapter 7: Cubes and Cube Roots – All Exercise Solutions | CBSE Board
Chapter 7: Cubes and Cube Roots
Exercise 7.1
1. Which of the following numbers are not perfect cubes?
(i) 216 (ii) 128 (iii) 1000 (iv) 100 (v) 46656
Solution: (i) 216
By prime factorisation ,we have
The prime factor 2 and 3 are appeared in a group of three . So, 216 is a perfect cube .
(ii) 128
By prime factorisation ,we have
Since there is only one 2 in the prime factorisation . So, we need (2 × 2 =) 4 to make it a perfect cube .
Therefore, 128 is not a perfect cube .
(iii) 1000
By prime factorisation ,we have
The prime factor 2 and 5 are appeared in a group of three . So, 1000 is a perfect cube .
(iv) 100
By prime factorisation ,we have
The prime factor 2 and 5 are not appeared in a group of three . So, 100 is not a perfect cube .
(v) 46656
By prime factorisation ,we have
The prime factor 2 and 3 are appeared in a group of three . So, 46656 is a perfect cube .
2. Find the smallest number by which each of the following numbers must be multiplied to obtain a perfect cube.
(i) 243 (ii) 256 (iii) 72 (iv) 675 (v) 100
Solution: (i) 243
We have , 
In this prime factorisation ,we find that there is no triplet of 3 . So, 243 is not a perfect cube .
To make it a perfect cube we multiply it by 3 .
Thus , 
is a perfect cube .
(ii) 256
We have, 
In this prime factorisation ,we find that there is no triplet of 2 . So, 256 is not a perfect cube .
To make it a perfect cube we multiply it by 2 .
Thus , 
is a perfect cube .
(iii) 72
We have ,
In this prime factorisation ,we find that there is no triplet of 2 . So, 256 is not a perfect cube .
To make it a perfect cube we multiply it by 2 .
Thus , is a perfect cube .
(iv) 675
We have , 
In this prime factorisation ,we find that there is no triplet of 5 . So, 675 is not a perfect cube .
To make it a perfect cube we multiply it by 5 .
Thus , 
is a perfect cube .
(v) 100
We have ,
In this prime factorisation ,we find that there is no triplet of 2 and 5 . So, 100 is not a perfect cube .
To make it a perfect cube we multiply it by 2 × 5 .
Thus , 
is a perfect cube .
3. Find the smallest number by which each of the following numbers must be divided to obtain a perfect cube.
(i) 81 (ii) 128 (iii) 135 (iv) 192 (v) 704
Solution: (i) 81
We have , 
The primes 3 do not appear in groups of three . So, 81 is not a perfect cube .
If we divide 81 by 3 , then the prime factorization of the quotient will not contain 3 . So, the resulting perfect cube is 
.
(ii) 128
We have , 
The primes 2 do not appear in groups of three . So, 128 is not a perfect cube .
If we divide 128 by 2 , then the prime factorization of the quotient will not contain 2 . So, the resulting perfect cube is 
.
(iii) 135
We have , 
The primes 5 do not appear in groups of three . So, 135 is not a perfect cube .
If we divide 135 by 5 , then the prime factorization of the quotient will not contain 5 . So, the resulting perfect cube is 
.
(iv) 192
We have, 
The primes 3 do not appear in groups of three . So, 192 is not a perfect cube .
If we divide 192 by 3 , then the prime factorization of the quotient will not contain 3 . So, the resulting perfect cube is 
.
(v) 704
We have , 
The primes 11 do not appear in groups of three . So, 81 is not a perfect cube .
If we divide 81 by 11 , then the prime factorization of the quotient will not contain 11 . So, the resulting perfect cube is 
.
4. Parikshit makes a cuboid of plasticine of sides 5 cm, 2 cm, 5 cm. How many such cuboids will he need to form a cube?
Solution: Given , the sides of the cuboid are 5 cm , 2 cm and 5 cm .
We have , 
In this prime factorisation ,we find that there is no triplet of 2 and 5 . So, 50 is not a perfect cube .
To make it a perfect cube we multiply it by 2 × 2 × 5 .
Thus , 
is a perfect cube .
Hence , he needs 2×2×5 = 20 cuboids .
Exercise 7.2
1. Find the cube root of each of the following numbers by prime factorisation method.
(i) 64 (ii) 512 (iii) 10648 (iv) 27000 (v) 15625 (vi) 13824 (vii) 110592 (viii) 46656 (ix) 175616 (x) 91125
Solution: (i) 64
By prime factorization method , we have , 64 = 2×2×2×2×2×2
So, 
(ii) 512
By prime factorization method , we have , 512 = 2×2×2×2×2×2×2×2×2
So, 
(iii) 10648
By prime factorization method , we have,
10648 = 2 × 2 × 2 × 11 × 11 × 11
So, 
(iv) 27000
By prime factorization method , we have 27000 = 2×2×2×3×3×3×5×5×5
So, 
(v) 15625
By prime factorization method , we have 15625 = 5×5×5×5×5×5
So, 
(vi) 13824
By prime factorization method , we have
13824 = 2×2×2×2×2×2×2×2×2×3×3×3
So, 
(vii) 110592
By prime factorization method , we have
110592 = 2×2×2×2×2×2×2×2×2×2×2×2×3×3×3
So, 
(viii) 46656
By prime factorization method , we have
46656 = 2×2×2×2×2×2×3×3×3×3×3×3
So, 
(ix) 175616
By prime factorization method ,we have
175616 = 2×2×2×2×2×2×2×2×2×7×7×7
So, 
(x) 91125
By prime factorization method , we have
91125 = 3×3×3×3×3×3×5×5×5
So, 
2. State true or false.
(i) Cube of any odd number is even.
(ii) A perfect cube does not end with two zeros.
(iii) If square of a number ends with 5, then its cube ends with 25.
(iv) There is no perfect cube which ends with 8.
(v) The cube of a two digit number may be a three digit number.
(vi) The cube of a two digit number may have seven or more digits.
(vii) The cube of a single digit number may be a single digit number.
Answer: (i) Cube of any odd number is even.
False
[ The cube of an odd number is always odd.
Example: 3³ = 27 (odd) ]
(ii) A perfect cube does not end with two zeros.
True
[ If a perfect cube ends in zeros, the number of zeros must be a multiple of 3, not 2.]
(iii) If square of a number ends with 5, then its cube ends with 25.
False
[ Example: 5² = 25, but 5³ = 125, which ends in 125, not 25.]
(iv) There is no perfect cube which ends with 8.
False
[ 2³ = 8 , 12³ = 1728 , etc. So perfect cubes can end with 8.]
(v) The cube of a two digit number may be a three digit number.
False
[ The smallest two-digit number is 10, and 10³=1000, which is a four-digit number. So the cube of a two-digit number is at least 4 digits.]
(vi) The cube of a two digit number may have seven or more digits.
False
[ The largest two-digit number is 99, and 99³ = 970299, which is a six-digit number. So it cannot have seven or more digits.]
(vii) The cube of a single digit number may be a single digit number.
True
[ Example: 2³ = 8 , which is a single-digit number. ]
3. You are told that 1,331 is a perfect cube. Can you guess without factorisation what is its cube root? Similarly, guess the cube roots of 4913, 12167, 32768.
Solution: (a) 1331
We have , 1331
We get 1 is second group and 331 is first groups as two groups of one digit and three digits each .
The number 331 ends with 1 , We know that 1 comes at the unit’s place of a number only when it’s cube root ends in 1 . So, we get 1 at the unit’s place of the cube root .
For second group , i.e. , 1
We know that 1³ = 1
Therefore , 1 will be taken at the tens place of the required cube root .
So , 
.
(b) 4913
We have , 4913
We get 4 is second group and 913 is first groups as two groups of one digit and three digits each .
The number 913 ends with 3 , We know that 7 comes at the unit’s place of a number only when it’s cube root ends in 3 . So, we get 7 at the unit’s place of the cube root .
For second group , i.e. , 4
We know that 
and 
Also , 1 < 4 < 8
We take the one’s place of the smallest number 4 as the ten’s place of the required cube root . So, we get 
.
(b) 12167
We have , 12167
We get 12 is second group and 167 is first groups as two groups of one digit and three digits each .
The number 167 ends with 7 , We know that 3 comes at the unit’s place of a number only when it’s cube root ends in 7 . So, we get 3 at the unit’s place of the cube root .
For second group , i.e. , 12
We know that 
and 
Also , 8 < 12 < 27
We take the one’s place of the smallest number 2 as the ten’s place of the required cube root . So, we get 
.
(c) 32768
We have , 32768
We get 32 is second group and 768 is first groups as two groups of one digit and three digits each .
The number 768 ends with 2 , We know that 2 comes at the unit’s place of a number only when it’s cube root ends in 8 . So, we get 2 at the unit’s place of the cube root .
For second group , i.e. , 32
We know that 
and 
Also , 27 < 32 < 64
We take the one’s place of the smallest number 3 as the ten’s place of the required cube root . So, we get 
.
Extra Question :
(d) 592704
We have , 592704
We get 592 is second group and 704 is first groups as two groups of one digit and three digits each .
The number 704 ends with 4 , We know that 4 comes at the unit’s place of a number only when it’s cube root ends in 4 . So, we get 4 at the unit’s place of the cube root .
For second group , i.e. , 592
We know that 
and 
Also , 512 < 592 < 729
We take the one’s place of the smallest number 8 as the ten’s place of the required cube root . So, we get 
.
(e) 912673
We have , 912673
We get 912 is second group and 673 is first groups as two groups of one digit and three digits each .
The number 673 ends with 7 , We know that 7 comes at the unit’s place of a number only when it’s cube root ends in 3 . So, we get 7 at the unit’s place of the cube root .
For second group , i.e. , 673
We know that and
Also , 512 < 673 < 729
We take the one’s place of the smallest number 8 as the ten’s place of the required cube root . So, we get 
.
Posted 7 months ago