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Chapter 6: Square Roots and Cube Roots > Exercise 6.4

Exercise 6.4

1. Find the cube of the following numbers

(i) 8

Solution:

83 = × × =

(ii) 16

Solution:

163 = × × =

(iii) 21

Solution:

213 = × × =

(iv) 30

Solution:

303 = × × =

2. Test whether the given numbers are perfect cubes or not.

(i) 243

Solution:

243 is a perfect cube:

243 = × × × × .

Since the prime factors cannot be grouped into triplets, 243 is not a perfect cube.

(ii) 516

Solution:

516 is a perfect cube:

516 = × × × .

Since the prime factors cannot be grouped into triplets, 516 is not a perfect cube.

(iii) 729

Solution:

729 is a perfect cube:

729 = 3 × 3 × 3 × 3 × 3 × 3 = 3×33 = .

Therefore, 729 is a perfect cube.

(iv) 8000

Solution:

8000 is a perfect cube:

8000 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5 × 5 = .

Therefore, 8000 is a perfect cube.

(v) 2700

Solution:

2700 is a perfect cube:

2700 = × × × × × × .

Since the prime factors cannot be grouped into triplets, 2700 is not a perfect cube.

3. Find the smallest number by which 8788 must be multiplied to obtain a perfect cube?

Solution:

Prime factorize 8788: 8788 = ×

To make it a perfect cube, we need one more .

The smallest number to multiply is .

4. What smallest number should 7803 be multiplied with so that the product becomes a perfect cube?

Solution:

Prime factorize 7803: 7803 = ×

Where, 27 = × × and 289 = ×

= ×

To make it a perfect cube, we need one more .

The smallest number to multiply is .

5. Find the smallest number by which 8640 must be divided so that the quotient is a perfect cube?

Solution:

Prime factorize 8640: 8640 = × ×

To make it a perfect cube by division, we need to remove and .

The smallest number to divide by is 2 × 5 = .

6. Ravi made a cuboid of plasticine of dimensions 12cm, 8cm and 3cm. How many minimum number of such cuboids will be needed to form a cube?

Solution:

Volume of the cuboid = 12 × 8 × 3 = cm3

Prime factorize the dimensions: 12 = 22 × , 8 = 23, 3 = 31

To form a cube, each dimension needs to be a perfect cube. The LCM of the powers needs to be a multiple of .

We need one more , two more s.

The minimum number of cuboids needed is 2 × 3 × 3 = .

7. Find the smallest prime number dividing the sum 3¹¹ + 5¹³

Solution:

311 is . 513 is .

The sum of two odd numbers is .

The smallest prime number that divides any even number is .