Mathematics – Class 8 – Chapter 7 – Cubes and Cube Root – Exercise 7.2 – NCERT Exercise Solution

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

(i) 64

Solution:

Prime factorization of 64:

64 = 2×2×2×2×2×2

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

Here, 64 can be grouped into triplets of equal factors,

So, 64 = 2×2 = 4

Hence, 4 is cube root of 64.

(ii) 512

Solution:

Prime factorization of 512:

512 = 2×2×2×2×2×2×2×2×2

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

Here, 512 can be grouped into triplets of equal factors,

So, 512 = 2×2×2 = 8

Hence, 8 is cube root of 512.

(iii) 10648

Solution:

Prime factorization of 10648:

10648 = 2×2×2×11×11×11

Now, 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,

So, 10648 = 2 ×11 = 22

Hence, 22 is cube root of 10648.

(iv) 27000

Solution:

Prime factorization of 27000:

27000 = 2×2×2×3×3×3×3×5×5×5

Now, 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,

So, 27000 = (2×3×5) = 30

Hence, 30 is cube root of 27000.

(v) 15625

Solution:

Prime factorization of 15625:

15625 = 5×5×5×5×5×5

Now, 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,

So, 15625 = (5×5) = 25

Hence, 25 is cube root of 15625.

(vi) 13824

Solution:

Prime factorization of 13824:

13824 = 2×2×2×2×2×2×2×2×2×3×3×3

Now, 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,

So, 13824 = (2×2× 2×3) = 24

Hence, 24 is cube root of 13824.

(vii) 110592

Solution:

Prime factorization of 110592:

110592 = 2×2×2×2×2×2×2×2×2×2×2×2×3×3×3

Now, 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,

So, 110592 = (2×2×2×2 × 3) = 48

Hence, 48 is cube root of 110592.

(viii) 46656

Solution:

Prime factorization of 46656:

46656 = 2×2×2×2×2×2×3×3×3×3×3×3

Now, 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 triplets of equal factors,

So, 46656 = (2×2×3×3) = 36

Hence, 36 is cube root of 46656.

(ix) 175616

Solution:

Prime factorization of 175616:

175616 = 2×2×2×2×2×2×2×2×2×7×7×7

Now, 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 triplets of equal factors,

So, 175616 = (2×2×2×7) = 56

Hence, 56 is cube root of 175616.

(x) 91125

Solution:

Prime factorization of 91125:

91125 = 3×3×3×3×3×3×3×5×5×5

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

Here, 91125 can be grouped into triplets of equal factors,

So, 91125 = (3×3×5) = 45

Hence, 45 is 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 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

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:

(i)According to given question:

 By grouping the digits 1,331,

we get 1 and 331

We know that, if the unit digit of cube is 1, then the unit digit of cube root is 1.

So, We get 1 as unit digit of the cube root of 1331.

Here cube of 1 matches with the number of second group.

Then, The ten’s digit of our cube root is taken as the unit place of smallest number.

As we know that, the unit digit of the cube of a number having digit as unit place 1 is 1.

Hence, ∛1331 = 11

(ii) Now, for 4,913

By grouping the digits of 4,913,

we get 4 and 913

We know that, if the unit digit of cube is 3, then unit digit of cube root is 7.

So, we get 7 as unit digit of the cube root of 4913.

We know 1³ = 1 and 2³ = 8,

So, 1 > 4 > 8.

Now, 1 is taken as ten digit of cube root.

Hence, ∛4913 = 17

(iii) For 12,167

By grouping the digits of 12,167,

we get 12 and 167.

We know that, if the unit digit of cube is 7, then unit digit of cube root is 3.

So, 3 is the unit digit of the cube root of 12167

We know 2³ = 8 and 3³ = 27

So, 8 > 12 > 27.

Thus, 2 is taken as ten digit of cube root.

Hence, ∛12167= 23

(iv) For 32,768

By grouping the digits of 32,768,

we get 32 and 768.

We know that, if the unit digit of cube is 8, then unit digit of cube root is 2.

So, 2 is the unit digit of the cube root of 32768.

We know 3³ = 27 and 4³ = 64

So, 27 > 32 > 64.

Thus, 3 is taken as ten digit of cube root.

Hence, ∛32768= 32