Let $x$ be a root of $f=t^3-t^2+t+2 \in \mathbb{Q}[t]$ and $K=\mathbb{Q}(x)$. Express $\frac{1}{x-1}$ in the form $ax^2+bx+c$, where $a,b,c\in \mathbb{Q}$.
I have proved that $f$ is the minimal polynomial of $x$ over $\mathbb{Q}$ but I am stuck showing the above claim. I tried writing $\frac{1}{x-1}=ax^2+bx+c$ and solve for $a,b,c$ but it didn't seem to work. Any idea?
Using the Extended Euclidean Algorithm as implemented in this answer, we get $$ \begin{array}{r} &&x^2&1&-(x+2)/3\\\hline 1&0&1&-1&(1-x)/3\\ 0&1&-x^2&x^2+1&(x^3-x^2+x+2)/3\\ x^3-x^2+x+2&x-1&x+2&-3&0\\ \end{array} $$ which means that $$ \left(\vphantom{x^2}x-1\right)\left(x^2+1\right)+\left(x^3-x^2+x+2\right)\cdot\left(-1\vphantom{x^2}\right)=-3 $$ Therefore, $$ -\frac{x^2+1}3\equiv\frac1{x-1}\pmod{x^3-x^2+x+2} $$
We have $x^3-x^2+x-1+3=0$, so $(x-1)(x^2+1)=-3$, giving $$\frac{1}{x-1}=-\frac{1}{3}(x^2+1).$$
Remark: We got "lucky," since our cubic has a very nice shape. However, an uglier polynomial could still be written as a polynomial in $x-1$ (Taylor expansion). So the same idea works.
