Proof that $(n^7-n^3)(n^5+n^3)+n^{21}-n^{13}$ is a multiple of $3$. I proved that $$(n^7-n^3)(n^5+n^3)+n^{21}-n^{13}$$ is a multiple of $3$ through the use of Little Fermat's theorem but i want to know if there exist other proofs(maybe for induction). How can I demonstrate it?
This my proof:
$$n^3(n^4-1)(n^5+n^3)+n^{13}(n^8-1)$$
Now i know that $$n^4-1\equiv 0\pmod 3 (n\neq 3k)$$ and $$n^8-1\equiv 0\pmod 3 (n\neq 3k).$$
Therefore I proved that $(n^7-n^3)(n^5+n^3)+n^{21}-n^{13}$ is a multiple of $3$.

Another approach
we see that $n(n-1)(n+1)$ divides both $n^7-n^3=n^3(n-1)(n+1)(n^2+1)$ and $n^{21}-n^{13}=n^{13}(n-1)(n+1)(n^2+1)(n^4+1)$
and we know that $(n-1)n(n+1)$ is a multiple of $3$ because the product of three consecutive integers is divisible by $3$.
