Question

Question 4
For any positive integer n, prove that $n^3–n$ is divisible by 6.

Answer 1

Let $a=n^3-n$
$=n(n^2-1)=n(n+1)(n-1)$
This is the product of three consecutive positive integers. We know that the product of three consecutive positive integers is divisible by 2 as well as 3. So, it must be divisible by 6.