Prove that $6$ divides $n^3+11n$?

$$n^3+11n=\underbrace{(n-1)n(n+1)}_{\text{Product of three consecutive integers}}+12n$$

See The product of n consecutive integers is divisible by n factorial

OR The product of n consecutive integers is divisible by n! (without using the properties of binomial coefficients)

It is easier to understand the part for factor $3$ to check if $n^3+11n = 0 \pmod 3$.

Then either $n = 0 \pmod 3$ or $n = \pm 1 \pmod 3$.

The $n^2 + 11$ is then $0 \pmod 3$.

You can also prove this by induction.

$$(n+1)^3+11(n+1)=(n^3+11n) + 3n(n+1)$$

$6$ divides the first part by induction hypothesis. Then $3$ divides $3$ and $2$ divides $n(n+1)$ (either $n$ or $n+1$ is even) so $6$ divides $3n(n+1)$.