NCERT Solution Class 8th Maths Chapter – 6 Cubes and Cube Roots Exercise – 6.2

1. Find the cube root of each of the following numbers by prime factorisation method.

(i) 64

Solution: 

64 = 2×2×2×2×2×2
By grouping the factors in triplets of equal factors, 64 = (2×2×2)×(2×2×2)

Here, 64 can be grouped into triplets of equal factors,
∴ 64 = 2×2 = 4
Hence, 4 is cube root of 64.

(ii) 512

Solution: 

512 = 2×2×2×2×2×2×2×2×2
By grouping the factors in triplets of equal factors, 512 = (2×2×2)×(2×2×2)×(2×2×2)

Here, 512 can be grouped into triplets of equal factors,
∴ 512 = 2×2×2 = 8
Hence, 8 is cube root of 512.

(iii) 10648

Solution: 

10648 = 2×2×2×11×11×11
By grouping the factors in triplets of equal factors, 10648 = (2×2×2)×(11×11×11)

Here, 10648 can be grouped into triplets of equal factors,
∴ 10648 = 2 ×11 = 22
Hence, 22 is cube root of 10648.

(iv) 27000

Solution: 

27000 = 2×2×2×3×3×3×3×5×5×5
By grouping the factors in triplets of equal factors, 27000 = (2×2×2)×(3×3×3)×(5×5×5)

Here, 27000 can be grouped into triplets of equal factors,
∴ 27000 = (2×3×5) = 30
Hence, 30 is cube root of 27000.

(v) 15625

Solution: 

15625 = 5×5×5×5×5×5
By grouping the factors in triplets of equal factors, 15625 = (5×5×5)×(5×5×5)

Here, 15625 can be grouped into triplets of equal factors,
∴ 15625 = (5×5) = 25
Hence, 25 is cube root of 15625.

(vi) 13824

Solution: 

13824 = 2×2×2×2×2×2×2×2×2×3×3×3
By grouping the factors in triplets of equal factors,
13824 = (2×2×2)×(2×2×2)×(2×2×2)×(3×3×3)

Here, 13824 can be grouped into triplets of equal factors,
∴ 13824 = (2×2× 2×3) = 24
Hence, 24 is cube root of 13824.

(vii) 110592

Solution: 

110592 = 2×2×2×2×2×2×2×2×2×2×2×2×3×3×3
By grouping the factors in triplets of equal factors,
110592 = (2×2×2)×(2×2×2)×(2×2×2)×(2×2×2)×(3×3×3)

Here, 110592 can be grouped into triplets of equal factors,
∴ 110592 = (2×2×2×2 × 3) = 48
Hence, 48 is the cube root of 110592.

(viii) 46656

Solution:

46656 = 2×2×2×2×2×2×3×3×3×3×3×3
By grouping the factors in triplets of equal factors,
46656 = (2×2×2)×(2×2×2)×(3×3×3)×(3×3×3)

Here, 46656 can be grouped into triples of equal factors,
∴ 46656 = (2×2×3×3) = 36
Hence, 36 is the cube root of 46656.

(ix) 175616

Solution: 

175616 = 2×2×2×2×2×2×2×2×2×7×7×7
By grouping the factors in triplets of equal factors,
175616 = (2×2×2)×(2×2×2)×(2×2×2)×(7×7×7)

Here, 175616 can be grouped into triples of equal factors,
∴ 175616 = (2×2×2×7) = 56
Hence, 56 is the cube root of 175616.

(x) 91125

Solution: 

91125 = 3×3×3×3×3×3×3×5×5×5
By grouping the factors in triplets of equal factors, 91125 = (3×3×3)×(3×3×3)×(5×5×5)

Here, 91125 can be grouped into triples of equal factors,
∴ 91125 = (3×3×5) = 45
Hence, 45 is the cube root of 91125.

2. State true or false.

(i) Cube of any odd number is even.
Solution:  False

(ii) A perfect cube does not end with two zeros.
Solution: True

(iii) If the cube of a number ends with 5, then its cube ends with 25.
Solution: False

(iv) There is no perfect cube which ends with 8.
Solution: False

(v) The cube of a two-digit number may be a three-digit number.
Solution: False

(vi) The cube of a two-digit number may have seven or more digits.
Solution: False

(vii) The cube of a single-digit number may be a single-digit number.
Solution: True