looking for the inverse of $3+2\alpha^2$ over $\Bbb Q(\alpha)$ with $\alpha=\sqrt[3]{2}$ I have $\alpha=\sqrt[3]{2}$ and want to calculate the inverse of $3+2\alpha^2$ over $\Bbb Q(\alpha)$. There's a hint which tells me to look at the minimal polynomial $m_\alpha$ of $\alpha$ over $\Bbb Q$ an work in $\Bbb Q[X]/(m_\alpha)$. I know that this polynomial is $X^3-2$ but I don't know how to proceed.

If $R$ is any PID and $f \in R$ is an element, then division in $R/(f)$ works as follows: If $g \in R$ is coprime to $f$, then the extended Euclidean algorithm gives $u,v \in R$ with $uf + vg = 1$. This implies that $[v]$ is inverse to $[g]$ in $R/(f)$. You can apply this in particular if $R=K[X]$ for a field $K$, and when $f$ is some specific polynomial.
Here, there is also another method. We calculate in the Galois closure $\mathbb{Q}(\zeta_3,\sqrt[3]{2})$ of $\mathbb{Q}(\sqrt[3]{2})$. The conjugates of $\sqrt[3]{2}$ are $\zeta_3^i \sqrt[3]{2}$ with $i=0,1,2$. So we compute:
$$\frac{1}{3+2 \alpha^2} = \frac{(3+2 (\zeta_3 \alpha)^2) (3 + 2 (\zeta_3^2 \alpha)^2)}{(3+2 \alpha^2)(3+2 (\zeta_3 \alpha)^2) (3 + 2 (\zeta_3^2 \alpha)^2)}$$
$$=\frac{9+6 \zeta_3 \alpha^2+6 \zeta_3^2 \alpha^2 + 4 \zeta_3^3 \alpha^4}{(3+2 \alpha^2)(3+2 (\zeta_3 \alpha)^2) (3 + 2 (\zeta_3^2 \alpha)^2)}=\frac{9+8 \alpha - 6 \alpha^2}{(3+2 \alpha^2)(3+2 (\zeta_3 \alpha)^2) (3 + 2 (\zeta_3^2 \alpha)^2)}$$
$$=\frac{9+8 \alpha - 6 \alpha^2}{(3+2 \alpha^2)(9+8 \alpha - 6 \alpha^2)}=\frac{9+8 \alpha - 6 \alpha^2}{59}.$$
