$\forall \ n \in \mathbb{N}: 6\mid 5n^3+n$ Show  $\forall \ n \in \mathbb{N}:  6\mid 5n^3+n$
Indeed,
we've to show $5n^3+n=0\pmod 6 $
note that for  $n\in \{0,1,2,3,4,5\}$ we've: $n^{3}=n\pmod 6$ then 
$5n^{3}=5n\pmod 6 \implies 5n^{3}+n=6n\pmod 6=0\pmod 6$ 
AM i right ?

Use induction
$$
5\Big(n+1\Big)^3 + \Big(n+1\Big) = 5 n^3 + n + 6 \Big( \tfrac{5}{2} n(n+1) +1 \Big)
$$
and as $n(n+1)$ is even we see that
$$
5\Big(n+1\Big)^3 + \Big(n+1\Big) = 5 n^3 + n + 6 k
$$
As it is true for $n=1$, it is true for all $n \ge 1$.
