Show that the difference between any integer and its cube is always divisible by 6 Show that the difference between any integer and its cube is always divisible by 6
 A: $$\begin{aligned}
\text{Diff}&=x^3-x \\
&=x(x^2-1) \\
&=x(x+1)(x-1) \qquad \text{via }(a^2-b^2)=(a+b)(a-b)
\end{aligned}$$

$=(x-1)\cdot x\cdot (x+1)$

So that always turns out to be product of three consecutive integers...which will always have one even number as 2*m and one number divisible by 3 as 3*n. Hence the product will always have minimal factors as 2 and 3...So, it is divisible by 6.
That's it!!!
A: $$
x^3-x=(x-1)\cdot x\cdot (x+1).
$$
Between three consecutive numbers, at least one is even, and at least one is multiple of $3$.
A: $$x-x^3=x(1-x^2)=x(1-x)(1+x)=-(x-1)x(x+1)$$
since $x-1,x,x+1$ are three consecutive numbers and their product is divisible by 6.
A: $$n^3-n=(n+1) n (n-1)= 6\binom{n+1}{3}$$
A: For the six first $n$, we get $n^3-n=0,0,6,24,60,120$.
Then $(n+6k)^3-(n+6k)=n^3-n+6m$.
A: If you factorize $n^3 − n$ you will get $(n-1)\cdot n\cdot (n+1)$ and you should know that product of $m$ consecutive integers is divisible by $m!$. (See here The product of n consecutive integers is divisible by n! (without using the properties of binomial coefficients))
Since $(n-1)\cdot n\cdot (n+1)$ is product of $3$ consecutive integers thus it will be divisible by $3!=6$.
