<s>[INST] Prove that $6$ divides $n^3+11n$? How can i show that
$$6\mid (n^3+11n)$$
My thoughts:
I show that
$$2\mid (n^3+11n)$$
$$3\mid (n^3+11n)$$
And
$$n^3+11n=n\cdot (n^2+11)$$
And if $n=x\cdot 3$ for all $x \in \mathbb{N}$ then:
$$3\mid (n^3+11n)$$
And if not:
The cross sum of$$n^2+11$$
is multiple of 3.
Can this be right or is there a simple trick?
 [/INST] $$
n^3+11n
= 12 \binom{n}{1} + 6 \binom{n}{2} + 6 \binom{n}{3}
= 6\left(2 \binom{n}{1} + \binom{n}{2} + \binom{n}{3}\right)
$$
This is obtained by using repeated differences and Newton's interpolation formula:
$$
\begin{array}{llll}
0 & 12 & 30 & 60 & 108 & \\
12 & 18 & 30 & 48 & \\
6 & 12 & 18 & \\
6 & 6 & \\
0 & \\
\end{array}
$$
</s>