Write it as an element of this ring? Since the degree of the irreducible polynomial $x^3+2x+2$ over $\mathbb{Q}[x]$ is odd, it has a real solution , let $a$. I am asked to express $\displaystyle{\frac{1}{1-a}}$ as an element of $\mathbb{Q}[a]$.
I have done the following:
Since $a$ is a real solution of $x^3+2x+2$, we have that $a^3+2a+2=0$.
$\frac{1}{1-a}=y \Rightarrow a=\frac{y-1}{y}$
$$a^3+2a+2=0 \\ \Rightarrow \frac{(y-1)^3}{y^3}+2\frac{y-1}{y}+2=0 \\ \Rightarrow (y-1)^3+2(y-1)y^2+2y^3=0 \\ \Rightarrow 5y^3-5y^2+3y-1=0 \\ \Rightarrow 5\left ( \frac{a-1}{a} \right )^3-5\left ( \frac{a-1}{a} \right )^2+3\left ( \frac{a-1}{a} \right )-1=0 $$
That what I have found is not a polynomial of $a$, with coefficients at $\mathbb{Q}$, right??
Is there an other way to write $\displaystyle{\frac{1}{1-a}}$ as an element of $\mathbb{Q}[a]$ ??

Without any trick: you know that $(1-a)^{-1}=\alpha+\beta a+\gamma a^2$ for some $\alpha,\beta,\gamma\in\mathbb{Q}$, that is
$$
(1-a)(\alpha+\beta a+\gamma a^2)=1
$$
This becomes
$$
\alpha+\beta a+\gamma a^2-\alpha a-\beta a^2-\gamma a^3=1
$$
Since $a^3=-2a-2$ this becomes
$$
(\alpha+2\gamma)+(\beta-\alpha+2\gamma)a+(\gamma-\beta)a^2=1
$$
or
\begin{cases}
\alpha+2\gamma=1\\
-\alpha+\beta+2\gamma=0\\
-\beta+\gamma=0
\end{cases}
This should be easy to solve.
