# For $m$ cubefree, $k^{6}|27m^{2}\Rightarrow k=1$ or $3$

For $m$ cubefree and $k$ integer $k^{6}|27m^{2}\Rightarrow k=1$ or $3$.

The "1" makes sense from $k^{6}|27m^{2}\Rightarrow ak^{3}=3m$ but not the 3.

So by assuming $9|m$

Any hints?

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Hint: Note that both $k^6$ and $27$ are perfect cubes; if $p$ is a prime divisor of $k$ that is not $3$, try concluding that $p^3 | m$.
@TKM Right: $k$ is a multiple of $3$ (or is $1$). If $k$ is divisible by $9$, then $k^6$ is divisible by an awfully large power of $3$, so $m$.... – user61527 Feb 27 '14 at 5:53