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?

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. 
