The DAV Class 8 Maths Book Solutions and **DAV Class 8 Maths Chapter 2 Worksheet 1** Solutions of Cubes and Cube Roots offer comprehensive answers to textbook questions.

## DAV Class 8 Maths Ch 2 WS 1 Solutions

Question 1.

Find the cubes of:

(i) 8

(ii) 13

(iii) 17

(iv) 1.3

(v) 0.06

(vi) 0.4

(vii) \(\frac{2}{3}\)

(viii) – 7

(ix) – 9

(x) – 12

Solution:

(i) 8^{3} = 8 × 8 × 8 = 512

(ii) 13^{3} = 13 × 13 × 13 = 2197

(iii) 17^{3} = 17 × 17 × 17 = 4913

(iv) 1.3^{3} = 1.3 × 1.3 × 1.3 = 2.197

(v) 0.06^{3} = 0.000216

(vi) (0.4)^{2} = 0.064

(vii) \(\left(\frac{2}{3}\right)^3=\frac{2 \times 2 \times 2}{3 \times 3 \times 3}=\frac{8}{27}\)

(viii) (- 7)^{3} = (- 7) × (- 7) × (- 7) = – 343

(ix) (- 9)^{3} = (- 9) × (- 9) × (- 9) = – 729

(x) (- 12)^{3} = (- 12) × (- 12) × (- 12) = – 1728.

Question 2.

Which of the following numbers are perfect cubes?

(i) 4096

(ii) 108

(iii) 392

(iv) – 27000

(v) \(\frac{-64}{1331}\)

Solution:

(i) 4096

∴ 4096 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

= 2^{3} × 2^{3} × 2^{3}

∴ 4096 is a perfect cube.

(ii) 108

108 = 2 × 2 × 3 × 3 × 3

= 2 × 2 × 3^{3}

Here 2 × 2 is left without triplet.

∴ 108 is not a perfect cube.

(iii) 392

392 = 2 × 2 × 2 × 7 × 7

= 2^{3} × 7 × 7

Here 7 × 7 is left without triplet.

∴ 392 is not a perfect cube.

(iv) – 27000 = (- 2) × (- 2) × (- 2) × (- 3) × (- 3) × (- 3) × (- 5) × (- 5) × (- 5)

= (- 2) × (- 3)^{3} × (- 5)

Here all have triplets.

∴ – 27000 is a perfect cube.

(v) \(\frac{-64}{1331}\)

⇒ – 64 = (- 4) × (- 4) × (- 4)

and 1331 = 11 × 11 × 11

∴ \(\frac{-64}{1331}\) is a perfect square.

Question 3.

Find the smallest number by which 2560 must be multiplied so that the product is a perfect cube.

Solution:

∴ 2560 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 5

Here 5 is left out without triplet.

So for this triplet 5 × 5 is needed

∴ 5 × 5 = 25 is the smallest number by which the given 2560 number must be multiplied to make it a perfect cube.

Question 4.

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

Solution:

∴ 8788 = 2 × 2 × 13 × 13 × 13

Here 2 × 2 is left out for making a triplet

∴ 2 × 2 = 4 is the smallest number by which the given number must be divided to make it a perfect cube.

Question 5.

Write True or False for the following statements.

(i) 650 is not a perfect cube.

Solution:

True

(ii) Perfect cubes may end with two zeros.

Solution:

False

(iii) Perfect cubes of odd numbers may not always be odd.

Solution:

False

(iv) Cube of negative numbers are negative.

Solution:

True

(v) For a number to be a perfect cube, it must have prime factors in triplets.

Solution:

False

Additional Questions:

Question 1.

Using the following matter, fill in the blanks:

(i) 1 = 1 = 1^{3}

(ii) 3 + 5 = 8 = 2^{2}

(iii) 7 + 19 + 11 = ____ = ____

(iv) 13 + 15 + 17 + 19 = ____ = ____

(v) ____ = 125 = 5^{3}

Sol.

(iii) 7 + 9 + 11 = 27 = 3^{3}

(iv) 13 + 15 + 17 + 19 = 64 = 4^{3}

(u) 21 + 23 + 25 + 27 + 29 = 125 = 5^{3}

Question 2.

Using the given matter, fill in the blanks:

(i) 2^{3} – 1^{3} = 1 + 2 × 1 × 3 = 7

(ii) 3^{3} – 2^{3} = 1 + 3 × 2 × 3 = 19

(iii) 4^{3} – 3^{3} = ____ = ____

(iv) 5^{3} – 4^{3} = ____ = ____

(v) 12^{3} – 11^{3} = ____ = ____

(vi) 20^{3} – 19^{3} = ____ = ____

Solution:

(iii) 4^{3} – 3^{3} = 1 + 4 × 3 × 3 = 37

(iv) 5^{3} – 4^{3} = 1 + 5 × 4 × 3 = 61

(v) 12^{3} – 11^{3} = 1 + 12 × 11 × 3 = 397

(vi) 20^{3} – 19^{3} = 1 + 20 × 19 × 3 = 1141

Question 3.

Which of the following are perfect cubes? Also find their cube roots.

(i) 6859

(ii) 2025

(iii) 15625

(iv) 3375

Solution:

(i) ∴ 6859 = 19 × 19 × 19

= \(\sqrt{19^3}\) = 19

Here 6859 is perfect cube and its cube root is 19.

(ii) 2025 = 3 × 3 × 3 × 3 × 5 × 5

Here 3 and 5 are ungrouped of triplet.

Hence, 2025 is not a perfect cube.

(iii) 15625 = 5 × 5 × 5 × 5 × 5 × 5

= 5^{3} × 5^{3}

∴ 15625 is a perfect cube.

Hence, its cube root is 5 × 5 = 25

(iv) 3375 = 3 × 3 × 3 × 5 × 5

= 3^{3} × 5^{3}

∴ 3375 is a perfect cube.

Hence, its cube root is 3 × 5 = 15.

Question 4.

Find the cube root of 17576 by estimation method.

Solution:

The following steps are taken to proceed:

(i) Form groups of three, starting from the rightmost digit of 17576.

(ii) 17 576, here 576 has 3 digits whereas 17 has only two digits.

(iii) The digit 6 is at ones place. So we take the one’s place of the required cube root as 6.

(iv) Take the other group, i.e. 17. Cube of 2 is 8 and cube of 3 is 27

∴ 23 < 17 < 33

∴ The ten’s place is 2.

Hence the required cube root is 26.

Question 5.

Find the cube root of 13824 by prime factorisation method.

Solution:

13824 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3

= 2^{3} × 2^{3} × 2^{3} × 3^{3}

∴ \(\sqrt[3]{13824}\) = 2 × 2 × 2 × 3 = 24.

### DAV Class 8 Maths Chapter 2 Worksheet 1 Notes

The cube of a number is obtained by multiplying itself three times and is read as, the number raised to the power 3.

e.g. cube of 4 = 4^{3}

= 4 × 4 × 4 = 64

cube of 7 = 7^{3}

= 7 × 7 × 7 = 343

cube of 9 = 9^{3}

= 9 × 9 × 9 = 729

Here 64, 343 and 729 are called perfect cubes.

**Properties of Cubes of Numbers**

1. The cube of any even number is even,

e.g. 4^{3} = 64 (even),

8^{3} = 512 (even)

2. The cube of any odd number is odd,

e.g. (3)^{3} = 27 (odd),

(9)^{3} = 729 (odd)

3. The cube of any number multiple of 2 is divisible by 8,

e.g. (6)^{3} = 216 (divisible by 8)

(8)^{3} = 512 (divisible by 8)

(12)^{3} = 1728 (divisible by 8)

4. The cube of a negative number is always negative,

e.g. (- 2) = – 2 × – 2 × – 2 = – 8

(- 5)3 = – 5 × – 5 × – 5 = – 125

5. The cube of a positive number is always positive,

e.g. (3)^{3}= 27 (positive)

(4)^{3} = 64 (positive)

6. The cube of a rational number is equal to the cube of its numerator divided by the cube of its denominator,

e.g. \(\left(\frac{4}{5}\right)^3=\frac{4^3}{5^3}\)

= \(\frac{64}{125}\)

\(\left(\frac{3}{4}\right)^3=\frac{3^3}{4^3}\)

= \(\frac{27}{64}\)

Example 1.

Is 512 a perfect cube? What is the number whose cube is 512?

Solution:

Resolving 512 into prime factors.

∴ 512 = \(\underline{2 \times 2 \times 2} \times \underline{2 \times 2 \times 2} \times \underline{2 \times 2 \times 2}\)

512 = 2^{3} × 2^{3} × 2^{3}

512 = (2 × 2 × 2)^{3} = (8)

Hence, 512 is a perfect cube and 8 is the number whose cube is 512.

Example 2.

Is 1512 a perfect cube?

Solution:

Resolving 1512 into prime factors.

1512 = \(2 \times 2 \times 2 \times 3 \times 3 \times 3\) × 7

1512 = 2^{3} × 2^{3} × 7

Here 7 is ungrouped.

Hence, 1512 is not a perfect cube.

Example 3.

Is 53240 a perfect cube? If not, then by which smallest natural number should 53240 be divided so that the quotient is a perfect cube.

Solution:

Resolving 53240 into prime factors.

53240 = \(2 \times 2 \times 2 \times 5 \times \underline{11 \times 11 \times 11}\)

= 2^{3} × 5 × 11^{3}

Here 5 is left ungrouped.

So 53240 is not a perfect cube.

∴ 53240 ÷ 5 = 10648

Hence, 10648 is a perfect cube.

Example 4.

Is 68600 a perfect cube? If not, find the smallest number by which 6860C must be multiplied to get a perfect cube.

Solution:

Let us resolve 68600 into prime factors.

∴ 68600 = 2 × 2 × 2 × 5 × 5 × 7 × 7 × 7

= 23 × 5 × 5 × 7

Here 5 × 5 is left without triplet.

∴ 68600 is not a perfect cube.

To make it a perfect cube, we multiply it by 5.

∴ 68600 × 5 = 2 × 2 × 2 × 5 × 5 × 5 × 7 × 7 × 7

= 343000 which is a perfect cube.

Hence, 5 is the number by which the given number must be multiplied 1 to make it a perfect cube.