Prove that $(n^3 - 1)n^3(n^3 + 1)$ is divisible by $504$ How to prove that $(n^3 - 1)n^3(n^3 + 1)$ is divisible by $504$? Factoring $504$ yields $2^3 \cdot 3^2 \cdot 7$. Divisibility by $8$ can be easily seen, because if $n$ is even then $8 | n^3$, else $(n^3 - 1)$ and $(n^3 + 1)$ are both even and one of these is divisible by $4$ so $8|(n^3 - 1)(n^3 + 1)$. I'm stuck at proving divisibility by $9$ and $7$

For $7$: $(n^3-1)(n^3+1)=n^6-1$ then apply Fermat's little theorem.
For $3^2$ : First, $(k-1)k(k+1)$ has to be divisible by $3$.
Second, $(n^3-1)\cdot(n^3+1)=(n-1)(n^2+n+1)\cdot (n+1)(n^2-n+1)$ which has the factor $(n^2-1)$ which is divisible by 3 by Fermat's LT.
