I really can't get a grasp on how to prove this, because if $x$ = $\sqrt[6] n$ for some $n$, then $x^2$ = $a$ and $x^3$ = $b$, with $a$ and $b$ being different natural numbers right? Any help?
[Math] a natural number that is both a perfect square and a perfect cube is a perfect sixth power
discrete mathematics
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Best Answer
Every positive integer $>1$ can be factorized into a product of prime numbers. If the number is a perfect square, the exponents of these primes are even. And if the number is a perfect cube, the exponents are multiple of three.
If both things happen for the same number, the exponents are multiple of $2$ and $3$, that is, multiple of $6$, so the number is a perfect sixth power.