<s>[INST] How to show $(n-1)^3n^3(n+1)^3$ is divisible by 7 and 9? Yeah it looks like a basic, really elementary question, but i'm having hard time with it.
First i tried to show that it's divisible by 9
$$(n-1)^3n^3(n+1)^3 = ((n+1)(n-1))^3n^3 = (n^2-1)^3n^3 = (n^3-n)^3$$
and using eulers theorem we know that 
$$[n^{\varphi(9)} \equiv 1 (mod \ 9)] = [n^6 \equiv 1 (mod \ 9)]$$
My doubt : can we do that? Cause $n$ and $9$ have to be coprime.
Is it right direction? I'd love some help on this, cause i never did tasks which asks for proving divisiblity of some polynomial. Cheers!
 [/INST] It's divisible by $9$ and in fact $27$, but not necessarily by $7$.
To see it is divisible by $27$, use the fact that that one of $n$, $n - 1$, $n - 2$ is divisible by $3$.
Your way
Expand out what you have:
$$
(n^3 - n)^3
= n^{9} - 3 n^{7} + 3n^5 - n^3
$$
If $n$ is divisible by $3$, you are done.
Otherwise, as you notice, we have $n^6 \equiv 1 \pmod 9$.
This implies $n^9 \equiv n^3$, so the above is
$$
\equiv n^3 - 3n^7 + 3n^5 - n^3 = -3(n^7 - n^5) \pmod 9
$$
Now since you have one factor of $3$ for sure, you consider $n^7 - n^5$ modulo $3$.
Euler's theorem gives $n^{\varphi(3)} = n^2 \equiv 1 \pmod 3$,
so $n^7 \equiv n^5$.
Therefore $n^7 - n^5 \equiv 0 \pmod 3$, and the expression in question is
$$
\equiv 0 \pmod 9.
$$
</s>