Test: Cubes And Cube Roots- 1 - UPSC 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

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

For example, 11*11*11=121*11
Where one’s digit depends on 1,

121*11=1331 where again one’s digit depends on 1.

Hence one’s digit of a perfect cube of a number depends on one’s digit.

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