Question

For any positive integer n, prove that n^3 - n is divisible by 6.
Give correct solution.​

Answer 1

Proof:

Factorize $n^3-n$.

$n^3-n=n\times (n^2-1)=(n-1)\times n\times (n+1)$

$n-1$, $n$, $n+1$ are consecutive integers.

At least one is an even number.

At least one is a multiple of 3.

∴ $(n-1)\times n\times (n+1)$ is a multiple of 6.

∴ Hence, it is divisible by 6.