1. ## Complex numbers question

Firstly, I apologise if this is in the wrong forum. I spent quite a bit of time looking through the different forums and this seems to be the one most appropriate.

Let $\alpha$ be such that $\alpha^3 = -1$ and $\alpha\neq-1$. Evaluate $(\alpha^2(\alpha - 1)^2)^{-1}$ without calculating explicitly the values of $\alpha$.

So far I have rewritten $(\alpha^2(\alpha - 1)^2)^{-1}$ as $(\alpha^2-\alpha+2)^{-1}$ by substituting $\alpha^3 = -1$ but am not sure how to proceed from here.

2. ## Re: Complex numbers question

Yes, you are correct that $\alpha^2(\alpha- 1)^2= \alpha^2(\alpha^2- 2\alpha+ 1)$ $= \alpha^4- 2\alpha^3+ \alpha^2= -\alpha+ 2+ \alpha^2$ because $\alpha^3= 1$.

The fact that $\alpha= -1$ satisfies $\alpha^3= -1$ means that $\alpha^3+ 1$ factors:
$\alpha^3+ 1= (\alpha+ 1)(\alpha^2- \alpha+ 1)$. Here, since $\alpha^3+ 1= 0$ but $\alpha\ne 0$, we have $\alpha^2- \alpha+ 1= 0$.

Finally, that $\alpha^2- \alpha + 2= (\alpha^2- \alpha+ 1)+ 1$. Easy to find the reciprocal of that, isn't it?