When looking at this problem, we know that the cubes are the same, so we're going to be able to multiply the insides X -, 1, which is a binomial times X ^2 + X + 1, which is a trinomial. Now when we actually multiply those together, right now it's a monomial because it's one thing times another thing. But that's one of our formulas, and if we didn't remember it, we could actually go ahead and distribute and we'd get X ^3 + X ^2 + X -, X ^2 -, X -, 1, and the X squareds in the X's would cancel and we get the cube root of X ^3 -, 1. Now, that's not a monomial, so that's actually the farthest we could go. If it had been the cube root of X -1 ^3, then we could take the cube root in the cube and get X - 1. But it's not. It's an X ^3 - 1, not a monomial. Right now it's 2 terms. If it had been a single term, then we could take the cube root of the cube X - 1 ^3. Remember, means X - 1 * X - 1 * X - 1, which would give us X ^3 minus three X ^2 + 3 X -1 if I factored it all the way out.