Question
Which of the following are perfect cube?1728

Answer

On factorising 1728 into prime factors, we get
1728 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3
Group the factors in triples of equal factors as:
1728 = {2 × 2 × 2} × {2 × 2 × 2} × {3 × 3 × 3}
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.

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