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

Given:

$n^3\ -\ n$

To prove:

Here we have to prove that for any positive integer n, prove that $n^3-n$ is divisible by 6.

Solution:

Let us consider that: 

$x\ =\ n^3\ –\ n$

Taking n common:

$x\ =\ n(n^2\ –\ 1)$

$x\ =\ n(n^2\ –\ 1^2)$

Using property {$a^2-b^2=$ (a $+$ b)(a $-$ b)}:

$x\ =\ n(n\ +\ 1)(n\ –\ 1)$

We know that (n $-$ 1), (n) and (n $+$ 1) are three consecutive numbers. So we can conclude that:

• One number must be even, and $x$ is divisible by 2.

• One number must be multiple of 3, and $x$ is divisible by 3 also.

Now, 

If a number is divisible by 2 and 3 both then that number is divisible by 6.

So, $n^3\ –\ n$ is divisible by 6 for any positive integer n.