Question

Without using a calculator prove whether 1728 is a perfect cube

Answer 1

`^3\sqrt[]{1728}^{}^{}^{}`
`^3\sqrt[]{(2\cdot2\cdot2)(2\cdot2\cdot2)(3\cdot3\cdot3)}`

Since the prime factors of 1728 can be grouped into triples of equal factors, it is a perfect cube.