How can i work out the following in factorials? : n(n^2-1) ?

Answer 1

Do you mean factorials or factors?
In factorials that would be
(n+1)!/(n-2)! or P(n+1, n-2)

In factors that would be n(n+1)(n-1)
If tihs is supposed to be set equal to zero and solved, n would be 0, 1, or -1

Answer 2

The strategy is to factor n^2-1 into (n+1)(n-1).

If you are looking for n(n^2-1)! you have
(n+1)!(n!)(n-1)!

Write out a few factors and you should see what this is.