Prove that $x^3 \equiv x \bmod 6$ for all integers $x$ 
Prove that $x^3 \equiv x \bmod 6$ for all integers $x$

I think I got it, but is this proof correct?
We can write any integer x in the form: $x = 6k, x = 6k + 1, x = 6k + 2, x = 6k + 3, x = 6k + 4$, and $x = 6k + 5$. 
If $x = 6k$, then $x^3 = 216k^3$. Then $x^3 - x = 216k^3 - 6k = 6(36k^3 - k)$. Thus, $6 | (x^3 - x)$. Thus, $x^3 \equiv x \pmod 6$. 
If $x = 6k + 1$. Then $x^3 - x = (216k^3 + 108k^2 + 18k + 1) - (6k + 1) = 6(36k^3 + 18k^2 + 2k)$ and so $6 | (x^3 - x)$ and $x^3 \equiv x \pmod 6$. 
If $x = 6k + 2$, then $x^3 - x = (216k^3 + 216k^2 + 72k + 8) - (6k + 2) = 6(36k^3 + 36k^2 + 11k + 1)$ and so $6 | (x^3 - x)$ and $x^3 \equiv x \pmod 6$ 
If $x = 6k + 3$, then $x^3 - x = (216k^3 + 324k^2 + 162k + 27) - (6k + 3) = 6(36k^3 + 54k^2 + 26 + 4)$ and so $6 | (x^3 - x)$ and $x^3 \equiv x \pmod 6 $
If $x = 6k + 4$, then $x^3 - x = (216k^3 + 432k^2 + 288k + 64) - (6k + 4) = 6(36k^3 + 72k + 47k + 10)$ and so $6 | (x^3 - x)$ and $x^3 \equiv x \pmod 6$.
If $x = 6k + 5$, then $x^3 - x = (216k^3 + 540k^2 + 450k + 125) - (6k + 5) = 6(36k^3 + 90k + 74k + 20)$ and so $6 | (x^3 - x)$ and $x^3 \equiv x \pmod 6$. 
In all cases, we have shown that $x^3 \equiv x \pmod 6$. QED. 

One has
$$x^3-x=x(x+1)(x-1)\qquad\forall x\ .$$
When $x$ is an integer then  at least one factor on the right is even, and exactly one factor on the right is divisible by $3$. It follows that for any $x\in{\mathbb Z}$ the right hand side is divisible by $6$, and so is the left hand side. That is to say: $x^3=x$ mod $6$ for all integers $x$.
