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!
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 $$ \equiv 0 \pmod 9. $$