Question

for any positive n , prove that n^3-n is divisible by 6.
please answer quickly it's urgent please please please please

Answer 1

n^3-n = n(n^2 - 1^2) = (n-1)n(n+1)..
clearly,these are concecutive 3 numbers..atleast one of them always has to be divisible by 3(since it is given that it is a +ve integer..)
we also know that among two concecutive numbers,atleast one of them is even and hence divisible by 2..
hence the number is divisible by both 3 and 2..
hence it is divisible by 6..