Question

Which of the following number is  not perfect cubes? 

1728

Answer 1

On factorising 1728 into prime factors, we get:

\[1728 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3\]

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

\[1728 = \left\{ 2 \times 2 \times 2 \right\} \times \left\{ 2 \times 2 \times 2 \right\} \times \left\{ 3 \times 3 \times 3 \right\}\]

It is evident that the prime factors of 1728 can be grouped into triples of equal factors and no factor is left over. Therefore, 1728 is a perfect cube.