Show steps to convert $C^3-3CS^2$ to $4C^3-3C$

Question

In Visual Complex Analysis by Needham, on page 16, he converts $C^3-3CS^2$ to $4C^3-3C$ where C and S represent $\cos(\theta)$ and $\sin(\theta)$ respectively. He does this using $C^2+S^2=1$. I have not been able to find the steps to go between these, even though it is trivial to plug in numbers and show that they are the same. Can you show me the algebra to equate these trig forms ?

I have tried factoring $C^2 + S^2$ out of $C^3-3CS^2$ but I just get $C$ remainder $-4CS^2$, which is to say that $C^2+S^2$ is not directly a factor. I can see that somehow an extra $3C^3$ got added, maybe by some convenient form of multiplication by 1/1, but I don't see it.

Answer 1

Recall: $\sin^2(x) = 1 - \cos^2(x)$

So, $C^3 - 3CS^2 = C^3 - 3C(1 - C^2) = C^3 - 3C + 3C^3 = 4C^3 - 3C$

Answer 2

$$C^3-3CS^2 = C^3 - 3C(1-C^2) = C^3 - 3C + 3C^3 = 4C^3-3C$$