## Cubes and Cube Roots Class 8 Extra Questions Maths Chapter 7

**Extra Questions for Class 8 Maths Chapter 7 Cubes and Cube Roots**

### Cubes and Cube Roots Class 8 Extra Questions Very Short Answer Type

Question 1.

Find the cubes of the following:

(a) 12

(b) -6

(c)

(d)

Solution:

Question 2.

Find the cubes of the following:

(a) 0.3

(b) 0.8

(c) .001

(d) 2 – 0.3

Sol.

(a) (0.3)^{3} = 0.3 × 0.3 × 0.3 = 0.027

(b) (0.8)^{3} = 0.8 × 0.8 × 0.8 = 0.512

(c) (0.001)^{3} = (0.001) × (0.001) × (0.001) = 0.000000001

(d) (2 – 0.3)^{3} = (1.7)^{3} = 1.7 × 1.7 × 1.7 = 4.913

Question 3.

Is 135 a perfect cube?

Solution:

Prime factorisation of 135, is:

135 = 3 × 3 × 3 × 5

We find that on making triplet, the number 5 does not make a group of the triplet.

Hence, 135 is not a perfect cube.

Question 4.

Find the cube roots of the following:

(a) 1728

(b) 3375

Solution:

Question 5.

Examine if (i) 200 (ii) 864 are perfect cubes.

Solution:

(i) 200 = 2 × 2 × 2 × 5 × 5

If we form triplet of equal factors, the number 2 forms a group of three whereas 5 does not do it.

Therefore, 200 is not a perfect cube.

(ii) We have 864 = 2 × 2 × 2 × 2 × 2

If we form triplet of equal factors, the number 2 and 3 form a group of three whereas another group of 2’s does not do so.

Therefore, 864 is not a perfect cube.

Question 6.

Find the smallest number by which 1323 may be multiplied so that the product is a perfect cube.

Solution:

1323 = 3 × 3 × 3 × 7 × 7

Since we required one more 7 to make a triplet of 7.

Therefore 7 is the smallest number by which 1323 may be multiplied to make it a perfect cube.

Question 7.

What is the smallest number by which 2916 should be divided so that the quotient is a perfect cube?

Solution:

Prime factorisation of

2916 = 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3

Since we required one more 2 to make a triplet

Therefore, the required smallest number by which 2916 should be divided to make it a perfect cube is 2 × 2 = 4, i.e., 2916 ÷ 4 = 729 which is a perfect cube.

Question 8.

Check whether 1728 is a perfect cube by using prime factorisation. (NCERT Exemplar)

Solution:

Prime factorisation of 1728 is

1728 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3

Since all prime factors can be grouped in triplets.

Therefore, 1728 is a perfect cube.

Question 9.

Using prime factorisation, find the cube root of 5832. (NCERT Exemplar)

Solution:

Question 10.

Solution:

### Cubes and Cube Roots Class 8 Extra Questions Short Answers Type

Question 11.

Solution:

Question 12.

Find the cube roots of

(i) 4

(ii) -0.729

Solution:

Question 13.

Express the following numbers as the sum of odd numbers using the given pattern

Solution:

Question 14.

Observe the following pattern and complete the blank spaces.

1^{3} = 1

Solution: