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.