The expression n3-n can be written as
n(n2-1) = n(n-1)(n+1)=(n-1)n(n+1).
Now for any positive integer n, either n will be even or n+1 will be even, so the above expression will have a factor 2 , as it has a even term. Similarly for any number n, n-1 and n+1 are three consecutive numbers so at least 1 of them has to be in the table of 3. ie, one of them has to have 3 as their factor.
Now since the expression has at least 3 and 2 as their factor, it means it is divisible by 6 always.