Express $1/(x-1)$ in the form $ax^2+bx+c$ 
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,
$$
-\frac{x^2+1}3\equiv\frac1{x-1}\pmod{x^3-x^2+x+2}
$$
