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

$$\begin{aligned}
\text{Diff}&=x^3-x \\
&=x(x^2-1) \\
&=x(x+1)(x-1) \qquad \text{via }(a^2-b^2)=(a+b)(a-b)
\end{aligned}$$

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

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!!!
