Question

Which of the following number is  not perfect cubes?

 64

Answer 1

On factorising 64 into prime factors, we get:

\[64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2\]

On grouping the factors in triples of equal factors, we get:

 

\[64 = \left\{ 2 \times 2 \times 2 \right\} \times \left\{ 2 \times 2 \times 2 \right\}\]

It is evident that the prime factors of 64 can be grouped into triples of equal factors and no factor is left over. Therefore, 64 is a perfect cube.