Let $R=\mathbb{Q}[x]$ with $x^3-x^2+x+2=0$. Find $(x-1)^{-2}$. I thought I could use the euclidean algorithm
$$x^3-x^2+x+2=(x-1)(x^2+1)+3$$
but I already have a constant after the first step. Also not sure that finding $(x-1)^{-1}$ would help anyway.
Well, $(x-1)^{-2}=((x-1)^{-1})^2$, no? As for the rest, you know that $$(x-1)(x^2+1)\equiv -3\pmod{x^3-x^2+x+2}$$ In other words, that $(x-1)\left(-\frac13x^2-\frac13\right)\equiv 1\pmod{x^3-x^2+x+2}$.