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

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)
Here is a proof by induction, 


setting $n=1$, one should get 
$$n^3+11n=1^3+11\cdot 1=12$$
obviously, the above number $12$ is divisible by $6$ hence statement is true for $n=1$.
Assume that the number $n^3+11n$ is divisible by $6$ for $n=k$ then 
$$k^3+11k=6\lambda$$or $$k^3=6\lambda-11k\tag 1$$
Now, setting $n=k+1$, $$(k+1)^3+11(k+1)$$$$=k^3+3k^2+14k+12$$
$$=6\lambda-11k+3k^2+14k+12$$
$$=6\lambda+3(k^2+k+4)$$
since, $k^2+k+4$ is even for all integer $k$ hence setting $k^2+k+4=2m$, 
$$=6\lambda+3(2m)$$
$$=6(\lambda+m)$$
since, $(\lambda +m)$ is an integer hence, the above number $6(\lambda+m)$ is divisible by $6$


Hence, $n^3+11n$ is divisible by $6$ for all integers $n\ge 1$