TITLE: Prove that $8640$ divides $n^9 - 6n^7 + 9n^5 - 4n^3$.
QUESTION [7 upvotes]: I found this problem in a book, I can't solve it unfortunately.
Prove that for all integer values $n$, $n^9 - 6n^7 + 9n^5 - 4n^3$ is divisible by $8640.$
So far I've noticed that $8460 = 6! \times 12$, also I've tried to simplify that expression and I've found that it's equal to this $n^3(n^3-3n-2)(n^3-3n+2)$, but I can't move on after that.

REPLY [7 votes]: $$
\begin{align}
&n^9-6n^7+9n^5-4n^3\\
&\small=362880\binom{n}{9}+1451520\binom{n}{8}+2298240\binom{n}{7}+1814400\binom{n}{6}\\
&\small+734400\binom{n}{5}+138240\binom{n}{4}+8640\binom{n}{3}\\
&=\small8640\left[42\binom{n}{9}+168\binom{n}{8}+266\binom{n}{7}+210\binom{n}{6}+85\binom{n}{5}+16\binom{n}{4}+\binom{n}{3}\right]
\end{align}
$$