Prove n^3-n is multiple of 6 for all n

factoring gives form n(n+1)(n-1) from here it should be realised that if n is not a multiple of 3 then n-1 is or n+1 is as multiples of 3 go up in 3s. Hence one of the values either n,n-1 or n+1 must be multiple of 3.