Question

For any positive integer n proove n^3-n is divisible by 6

Answer 1

look
${n}^{3} - n \: can \: be \: written \: as$
$n({n}^{2} - 1) = n( n + 1)(n - 1)$
so it gives us the product of 3 consecutive integers
like 17 18 19 90 91 92 since any 3 consecutive integers will involve an even no. and a number divisible by 3 so their product is divisible by 6