# RS Aggarwal Class 8 Math Fourth Chapter Cubes and Cube Roots Exercise 4A Solution

## EXERCISE 4A

**(1) Evaluate:**

(i) (8)^{3} = 8 × 8 × 8 = 512

(ii) (15)^{3} = 15 × 15 × 15 = 3375

(iii) (21)^{3} = 21 × 21 × 21 = 9261

(iv) (60)^{3} = 60 × 60 × 60 = 216000

**(2) Evaluate:**

(i) (1.2)^{3} = 1.2 × 1.2 × 1.2 = 1.728

(ii) (3.5)^{3} = 3.5 × 3.5 × 3.5 = 42.875

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

(iv) (0.05)^{3} = 0.05 × 0.05 × 0.05 = 0.000125

**(3) Evaluate:**

**(4) Which of the following numbers are perfect cubes? In case of perfect cube, find the number whose cube is the given number.**

(i) 125 = 5 × 5 × 5

(ii) 243 = 3 × 3 × 3 × 3 × 3

(iii) 343 = 7 × 7 × 7 = (7)^{3}; Thus, 343 is a perfect cube.

(iv) 256 = __2 × 2__ × __2 × 2__ × __2 × 2__ × __2 × 2__; Thus, 256 is not a perfect cube.

(v) 8000 = __2 × 2 × 2__ × __2 × 2 × 2__ × __5 × 5 × 5__

= (2 × 2 × 5)^{3} = (20)^{3}; Thus, 8000 is a perfect cube.

(vi) 9261 = __3 × 3 × 3__ × __7 × 7 × 7__

= (3 × 7)^{3} = 21^{3}; Thus, 9261 is a perfect cube.

(vii) 5324 = 2 × 2 × 11 × 11 × 11; Thus, 5324 is not a perfect cube.

(viii) 3375 = __3 × 3 × 3__ × __5 × 5 × 5__

= (3 × 5)^{3} = (15)^{3}; Thus, 3375 is a perfect cube.

**(5) Which of the following are the cubes of even numbers?**

Properties: The cubes of every even number are even.

(i) 216 = __2 × 2 × 2__ × __3 × 3 × 3__ = 2^{3} × 3^{3} = 6^{3}

(iii) 512 = __2 × 2 × 2__ × __2 × 2 × 2__ × __2 × 2 × 2__ = 2^{3} × 2^{3} × 2^{3} = (8)^{3}

(v) 1000 = __2 × 2 × 2__ × __5 × 5 × 5__ = 2^{3} × 5^{3} = (10)^{3}

**(6) Which of the following are the cubes of odd numbers?**

Properties: The cube of every odd number is odd.

(i) 125 = 5 × 5 × 5 = 5^{3}

(ii) 343 = 7 × 7 × 7 = 7^{3}

(v) 9261 = 3 × 3 × 3 × 7 × 7 × 7 = 3^{3} × 7^{3} = (21)^{3}

**(7) Find the smallest number by which 1323 must be multiplied so that the product is a perfect cube.**

Solution: Writing 1323 as a perfect of prime factors, we have:

1323 = __3 × 3 × 3__ × 7 × 7

Clearly, to make it a perfect cube, it must be multiplied by 7.

**(8) Find the smallest number by which 2560 must be divided so that the quotient is a perfect cube.**

Solution: Writing 2560 as a perfect of prime factors, we have:

2560 = __2 × 2 × 2__ × __2 × 2 × 2__ × __2 × 2 × 2__ × 5

Clearly, to make it a perfect cube, it must be multiplied by 5 × 5 = 25.

**(9) What is the smallest number by which 1600 must be divided so that the quotient is a perfect cube?**

Solution: Writing 1600 as a product of prime factors, we have:

1600 __= 2 × 2 × 2__ × __2 × 2 × 2__ × 5 × 5

Clearly, to make it a perfect cube, it must be divided by (5 × 5) = 25.

**(10) Find the smallest number by which 8788 be divided so that the quotient is a perfect cube.**

Solution: Writing 1600 as a product of prime factors, we have:

8788 = 2 × 2 × __13 × 13 × 13__

Clearly, to make it a perfect cube, it must be divided by (2 × 2) = 4.

Thank you so much sir