Test: Cubes And Cube Roots- 1 - Class 8 MCQ

Prime factorising 72, we get, 72 = 2 × 2 × 2 × 3 × 3. We know, a perfect cube has multiples of 3 as powers of prime factors. Here, number of 2's is 3 and number of 3's is 2. So we need to multiply another 3 in the factorization to make 72 a perfect cube.

Double of a here a + a = 2a
So cube of (2a)³ = 8a³

Prime factorisation of 27000,
is:
27000 = (2 x 2 x 2) x (3 x 3 x 3) x (5 x 5 x 5)
= 2³ x 3³ x 5³
∴ ∛27000
= 2 x 3 x 5
= 30
Thus, the cube root of 27000
= 300.

A natural number is said to be a perfect cube if it is the cube of some natural number.
Example
8 = 2 x 2 x 2
8 = 2 ³
8 is the perfect cube because it is a cube of 2 which is a natural number.
But 12 is not a perfect cube because it is not a cube of any natural numbers.

6 × 6 = 36 and 36 × 6 = 216, which is the cube of 6.

We know,

⇒ 31 + 33 + 35 + 37 + 39 + 41 = 216

And the cube of 6 is also 216

⇒ 6 × 6 × 6 = 216

1 × 1 × 1 = 1
2 × 2 × 2 = 8
3 × 3 × 3 = 27
4 × 4 × 4 = 64
5 × 5 × 5 = 125
Hence all are cubes

A number ending with 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 0 then it's cube ends with 1 , 8 , 7 , 4 , 5 , 6 , 3 , 2 , 0 respectively  

⇒ One's digit of 5832= 2
⇒ Cube of 2=2³ = 8 

So, the one's digit of cube of 5832=8

Solution:
The cube of an odd natural number is always odd. This is because when you multiply three odd numbers together (odd × odd × odd), the result is always odd.

Correct Answer: (b) odd