## Cubes and Cube Roots

**Question 1.****Find the cube root of each of the following numbers by prime factorisation method :(i)** 64

**(ii)**512

**(iii)**10648

**(iv)**27000

**(v)**15625

**(vi)**13824

**(vii)**110592

**(viii)**46656

**(ix)**175616

**(x)**91125

**Solution:****(i)** Resolving 64 into prime factors, we get

64 = 2 x 2 x 2 x 2 x 2 x 2

∴ **(ii)** Resolving 512 into prime factors, we get

512 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2

∴

**(iii)** Resolving 10648 into prime factors, we get

10648 = 2 x 2 x 2 x 11 x 11 x 11

∴

**(iv)** Resolving 27000 into prime factors, we get

27000 = 1000 x 27

= 10 x 10 x 10 x 3 x 3 x 3 = 2 x 5 x 2 x 5 x 2 x 5 x 3 x 3 x 3

= 2 x 2 x 2 x 3 x 3 x 3 x 5 x 5 x 5

∴

**(v)** Resolving 15625 into prime factors, we get

15625 = 5 x 5 x 5 x 5 x 5 x5

∴

**(vi) **Resolving 13824 into prime factors, we get

13824 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3

∴

**(vii)** Resolving 110592 in prime factors, we get

110592 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3

∴

**(viii)** Resolving 46656 in prime factors, we get

46656 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 x 3 x 3 x 3

∴

**(ix)** Resolving 175616 in prime factors, we get

175616 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 7 x 7 x 7 = (2 x 2 x 2 x 7)

= 56

**(x)** Resolving 91125 in prime factors, we get

91125 = 3 x 3 x 3 x 3 x 3 x 3 x 5 x 5 x 5

∴

**Question 2.****State true or false :****(i)** Cube of any odd number is even.**(ii)** A perfect cube does not end with two zeros.**(iii)** If square of a number ends with 5, then its cube ends with 25.**(iv)** There is no perfect cube which ends with 8.**(v)** The cube of a two digit number may be a three digit number.**(vi)** The cube of a two digit number may have seven or more digits.**(vii)** The cube of a single digit number may be a single digit number.

**Solution:****(i)** False**(ii)** True**(iii)** False, as 15^{2} = 225 and 15^{3} = 3375**(iv)** False, as 8 = 2^{3}, 1728 =12^{3}, etc.**(v)** False**(vi)** False, as 10^{3} = 1000, 99^{3} = 970299**(vii)** True, as 1^{3} = 1, 2^{3} =8.

**Question 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:****For 1331 :**

Units digit of the cube root of 1331 is 1 as unit’s digit of the cube root of numbers ending in 1 is 1 . After striking three digits from the right of 1331, we get the number 1. Since 1^{3} = 1, so the ten’s digit of the cube root of given number is 1.

∴

**For 4913 :**

Units digit of the cube root of 4913 is 7 as unit’s digit of cube root of numbers ending in 3 is 7. After striking three digits from the right of 4913, we get the number 4. As 1^{3} =1 and 2^{3} =8, so 1^{3} < 4 < 2^{3}. Therefore, the ten’s digit of cube root of 4913 is 1.

∴

**For 12167 :**

Unit’s digit of the cube root of 12167 is 3 as unit’s digit of cube root of numbers ending in 7 is 3. After striking three digits from the right of 12167, we get the number 12. As 2^{3} =8 and 3^{3} =27, so 2^{3} < 12 < 3^{3}. So, the ten’s digit of the cube root of 12167 is 2.

∴

**For 32768 :**

Unit’s digit of the cube root of 32768 is 2 as unit’s digit of cube root of numbers ending in 8 is 2. After striking three digits from the right of 32768, we get the number 32. As 3^{3} =27 and 4^{3} = 64, so 3^{3} < 32< 4^{3}. So, the ten’s digit of the cube root of 32768 is 3.

∴