Write it as an element of this ring?

Question

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 $\mathbb{Q}[a]$ ??

Answer 1

$$ a^3+2a+2=0\\ a^3+2a-3=-5\\ (a-1)(a^2+a+3)=-5\\ \frac{1}{5}(a^2+a+3)=\frac{1}{1-a} $$

Answer 2

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.