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

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

2. ## Re: Complex numbers question

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

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

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