Prove that $(n^3 - 1)n^3(n^3 + 1)$ is divisible by $504$

$(n^3 - 1)n^3(n^3 + 1) =n^3(n^6 -1)$

Since $\phi(7)=\phi(9)=6$, we get that $7$ and $9$ always divide $n^3(n^6 -1)$, by Euler–Fermat.

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.

For 9: if $n$ is divisible by 3, then the conclusion is obvious. If $n \equiv 1 \bmod 3$, then $n^3 \equiv 1 \bmod 9$ (proof: write $n^3 = (3k+1)^3 = 27k^3 + 27k^2 + 9k + 1$; all terms but the constant are killed modulo 9). If $n \equiv -1 \bmod 3$, then $n^3 \equiv -1 \bmod 9$ (proof is similar).

For 7, the conclusion follows from arberavdullahu's comment about cubic residues. If you'd rather it be more explicit: let $m \in \{0, \ldots, 6\}$ be such that $m \equiv n \bmod 7$; then $n^3 = (7k+m)^3 = 343k^3 + 147 k^2 m + 21km^2 + m^3$; all terms but $m^3$ die modulo 7, so you just have to check that all of $0^3, \ldots, 6^3$ are congruent to one of $-1, 0, 1$. In general, though, $n^p \equiv m^p \mod r$ as long as $n \equiv m \mod r$, so this step is easy (the mod-9 step, where we had to move from mod-3 considerations to mod-9, was a bit harder).