Expressing $(\alpha-1)^{-1}$ where $\alpha^3-\alpha^2+\alpha+2=0$ Let $E=\Bbb{Q}[\alpha]$ where $\alpha^3-\alpha^2+\alpha+2=0$
I would like to express $(\alpha^2+\alpha+1)(\alpha^2-1)$ and $(\alpha-1)^{-1}$ on the form $a+b\alpha+c\alpha^2$ where $(a,b,c)\in\Bbb{Q}^3.$
For the first one I get $-4\alpha-2$, but for the inverser I am not sure how can I do that, I would like to say that it's $\alpha-1$ but it seems 'silly'.

Let's try to represent $(\alpha - 1)^{-1}$ as $a + b\alpha + c\alpha^2$. We must have
$$ (\alpha - 1)(a + b\alpha + c\alpha^2) = a\alpha + b\alpha^2 + c\alpha^3 - a - b\alpha - c\alpha^2 \\
= c\alpha^3 + (b - c)\alpha^2 + (a-b)\alpha -a = \\
c(\alpha^3 - \alpha^2+\alpha +2) +b\alpha^2+(a-b-c)\alpha - (2c + a) = 1. $$
Thus, by comparing coefficients, we see that we want
$$ b = 0, a - b - c = 0, 2c + a = -1 $$
which is a linear system of equations whose solution is $b = 0$ and $a = c = -\frac{1}{3}$. That is, $(\alpha - 1)^{-1} = -\frac{1}{3} (1 + \alpha^2)$. Checking ourselves, we indeed have
$$ -\frac{1}{3}(\alpha - 1)(1 + \alpha^2) = -\frac{1}{3} \left( \alpha^3 - \alpha^2 + \alpha - 1 \right) = -\frac{1}{3} \left( \alpha^3 - \alpha^2 + \alpha + 2 - 3 \right) \\ 
= -\frac{1}{3}( -3) = 1.$$
