Compute $[(2+x^2)+k(x)(x^3-2)]^{-1}$ I have this exercise:

Let $I=<x^3-2>$ be ideal of $\mathbb{Q}[x]$. Consider the element
  $a=(x^2-2)+I\in\mathbb{Q}[x]/I$, compute $a^{-1}$.

I know that $\mathbb{Q}[x]/I$ is a field, and $a=(x^2-2)+k(x)(x^3-2)$ for some $k(x)\in\mathbb{Q}[x]$, then i want to have some $b\in\mathbb{Q}[x]$ such that $ab=ba=1$, but i haven't managed to know how to continue, can you help me please?
 A: Well, we do know that all elements can be reduced to the form $ax^2 + bx + c$, so that reduces our search space.
Now, we can actually just compute this directly, as we want
\begin{align*}
1 &= (ax^2 + bx + c)(x^2 - 2)\\
&= ax^4 + bx^3 + (c-2a)x^2 - 2bx - 2c\\
&= (c-2a)x^2 + (2a-2b)x + 2b-2c
\end{align*}
Thus we just have to solve
\begin{align*}
c-2a &= 0\\
a-b &= 0\\
2b-2c &= 1
\end{align*}
So we get $b=a=-1/2$ and $c=-1$
A: Subtract below equations, or use the extended Euclidean algorithm as in the $\rm\color{#0a0}{Example}$ below.
$\, \ \quad\qquad\qquad\qquad\begin{align} (x^2-2)(x^2+2) &\,=\, \color{#c00}2x-4\ \ \ {\rm by}\ \ (\color{#c00}{x^3})x = \color{#c00}2x\\
{\bf -}\qquad\qquad\quad\ x\,(x^2+2) &\,=\, \color{#c00}2+2x\ \ \ {\rm by}\ \ \ \color{#c00}{x^3 = 2}\\
\hline
\Rightarrow\ (x^2-x-2)(x^2+2) &\,=\, -6
\end{align}$
$\begin{align}{\rm generally}\ \ \ [\![1]\!]\,\qquad\qquad\quad (x^2-a)(x^2+a) &\,=\, \color{#c00}bx-a^2\ \ \, {\rm by}\ \ (\color{#c00}{x^3})x = \color{#c00}bx\\
[\![2]\!]\ \ \ \qquad\qquad\qquad\quad\ x\,(x^2+a) &\,=\, \color{#c00}b+ax\ \:\!\ \ \ {\rm by}\ \ \ \color{#c00}{x^3 = b}\\
a[\![1]\!]-b[\![2]\!]\Rightarrow\ (ax^2-bx-a^2)(x^2+a) &\,=\, -a^3-b^2
\end{align}$

$\rm\color{#0a0}{Example}$ Euclidean calculation from a different question. Exactly the same method works in the OP.
Generally it's easiest to use said augmented-matrix form  of the extended Euclidean algorithm, e.g. below we compute $\,1/g \pmod{\!f} = 1/(x^2\!+\!1) \pmod{\!x^3\!+\!2x\!+\!1}\,$ over $\,\Bbb Q,\,$ as in this answer.
$\,\begin{eqnarray}
[\![1]\!]&&  &&f = x^3\!+2x+1 &\!\!=&\, \left<\,\color{#c00}1,\,\color{#0a0}0\,\right>\quad\ \ \, {\rm i.e.}\ \qquad f\, =\ \color{#c00}1\cdot f\, +\, \color{#0a0}0\cdot g\\
[\![2]\!]&&  &&\qquad\ \,  g =x^2\!+1 &\!\!=&\, \left<\,\color{#c00}0,\,\color{#0a0}1\,\right>\quad\ \ \,{\rm i.e.}\ \qquad g\, =\ \color{#c00}0\cdot f\, +\, \color{#0a0}1\cdot g\\
[\![3]\!]&=&[\![1]\!]-x[\![2]\!]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! &&\qquad\qquad\ \ x+1 \,&\!\!=&\, \left<\,\color{#c00}1,\,\color{#0a0}{-x}\,\right>\ \ \ {\rm i.e.}\quad x\!+\!1\, =\, \color{#c00}1\cdot f\color{#0c0}{-\,x}\cdot g\\
[\![4]\!]&=&[\![2]\!]+(1\!-\!x)[\![3]\!]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! &&\qquad\qquad\qquad\ 2 \,&\!\!=&\, \left<\,\color{#c00}{1\!-\!x},\,\ \color{#0a0}{1\!-\!x+x^2}\,\right>\\
\end{eqnarray}$
Hence the prior line implies: $\ \ \ 2\  =\  (\color{#c00}{1\!-\!x})f + (\color{#0a0}{1\!-\!x\!+\!x^2})g $
$\!\!\bmod\! f\,$ this yields in $\Bbb Q[x]/f\!:\,\ 2\,  =\, (\color{#0a0}{1\!-\!x\!+\!x^2})g\ \Rightarrow\ {1/g= (\color{#0a0}{1\!-\!x\!+\!x^2})/2}$
