Prove that $6$ divides $n^3+11n$?
Solution 1:
$$n^3+11n=\underbrace{(n-1)n(n+1)}_{\text{Product of three consecutive integers}}+12n$$
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)
Solution 2:
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 $0 \pmod 3$.