Find the inverse element of an unknown complex root. I got asked if I could solve the following task: Let $f:=X^3-3X+4 \in \mathbb{Q}[X]$ be the polynomial with $\alpha \in \mathbb{C}$, $f(\alpha)=0$ and $K=\mathbb{Q}(\alpha)$. 
Furthermore, let $\beta := \alpha^2+\alpha+1$. 
Find the inverse of $\beta$. 
I tried solving it in a pretty naive way: 1=$\beta*\beta^{-1}=\beta^{-1}(\alpha^2+\alpha+1).$ 
Therefore if $\beta^{-1}*\alpha=0$ I should have $\beta^{-1}$. 
If $\beta^{-1}:=\alpha^2-3+4*\alpha^{-1}$ I should be done. 
However I don't think this is the correct answer, since the task further said, that $\beta^{-1}$ is of the form $\beta^{-1}=a_0+a_1\alpha+a_2\alpha^2$ for some $a_0,a_1,a_2 \in \mathbb{Q}$. 
Am i wrong? 
And if so, where did I go wrong? 
Thank you in advance!
 A: Note,
$$\beta = \alpha^2+\alpha+1=\frac{\alpha^3-1}{\alpha-1}
=\frac{3\alpha-5}{\alpha-1}=3-\frac2{\alpha-1}\tag 1$$
$$(\alpha -1)^3+3(\alpha -1)^2+2=0$$
Let $\gamma = \alpha -1 = re^{i\theta}$ be a complex root of the second equation and $b$ the real root. Then
$$b+\gamma + \bar{\gamma} = b+2r\cos\theta= -3$$
$$b(\gamma + \bar{\gamma}) + r^2= 2rb\cos\theta + r^2=0$$
$$br^2 = -2$$
which leads to 
$$r^6+6r^2-4=0,\>\>\>\>\>\cos\theta = \frac{r^3}4$$
and the solutions
$$r^2 = \sqrt[3]{2(\sqrt3+1)}- \sqrt[3]{2(\sqrt3-1)},\>\>\>\>\>\theta=\cos^{-1}\frac{r^3}4$$
Then, plug $\alpha -1 = re^{i\theta}$ into (1) to arrive at the inverse of $\beta$,
$$\beta^{-1}=\frac r{3r-2e^{-i\theta}}$$
A: Since $\alpha$ is a root of $x^3-3x+4$, we have that $\alpha^3-3\alpha+4=0$, so
$$\alpha^3=3\alpha-4.$$
Since, $\beta=\alpha^2+\alpha+1$ we have that $\beta\alpha=\alpha^3+\alpha^2+\alpha$. Hence
$$\beta\alpha=\alpha^2+4\alpha-4.$$
Similarly, $\beta\alpha^2=\alpha^3+4\alpha^2-4\alpha.$ Hence
$$\beta\alpha^2=4\alpha^2-\alpha-4.$$
Now we can find a system of linear equations to solve to find $\beta^{-1}$. We have that $1=\beta\left(a_0+a_1\alpha+a_2\alpha^2\right)$. So
$$1=a_0\beta+a_1\beta\alpha+a_2\beta\alpha^2=a_0\left(\alpha^2+\alpha+1\right)+a_1\left(\alpha^2+4\alpha-4\right)+a_2\left(4\alpha^2-\alpha-4\right).$$
Hence
$$1=\left(a_0+a_1+4a_2\right)\alpha^2+\left(a_0+4a_1-a_2\right)\alpha+\left(a_0-4a_1-4a_2\right).$$
Which gives us the following system:
$$a_0+a_1+4a_2=0$$
$$a_0+4a_1-a_2=0$$
$$a_0-4a_1-4a_2=1.$$
It follows that $a_0=\dfrac{17}{49}$, $a_1=\dfrac{-5}{49}$, and $a_2=\dfrac{-3}{49}$. So
$$\dfrac{1}{\alpha^2+\alpha+1}=\dfrac{17-5\alpha-3\alpha^2}{49}.$$
