Proof that $n^3 + 3n^2 + 2n$ is a multiple of $3$.
Solution 1:
Among three consecutive integers, one must be a multiple of three. Reason: if $n=3k$, we're done. If $n=3k+1$ then $n+2=3j$ is a multiple of three. If $n=3k+2$, then $n+1=3m$ is a multiple of three. In any case, $3\mid n(n+1)(n+2)$.
Solution 2:
HINT: $$n^3+3n^2+2n=n^3-n+3n^3+3n\equiv \underbrace{(n-1)n(n+1)}_{\text{ product of three consecutive integers}}\pmod3$$
Among three consecutive integers, on must be divisible by $3$