Without using a calculator prove whether 1728 is a perfect cube

Mllm asked Sep 17, 2023

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Mllm asked Sep 17, 2023

by Mllm

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10 votes

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

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

Simonwjackson answered Sep 23, 2023

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