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.
 A: As you said, the minimal polynomial is $m_\alpha(X)=X^3-2$.
Perform the Euclidean division of $m_\alpha(X)$ by $3+2X^2$. You'll find:
$$m_\alpha(X)=(3+2X^2)(aX+b) +c$$ with $a,b,c \in \mathbb{Q}$.
Now replace $X$ by $\alpha=\sqrt[3]{2}$. You'll find an interesting relation! 
I think I  made some mistakes in my answer. The way I wrote the division is wrong.
A: $$\alpha^3=2\ \Rightarrow\ \dfrac{\overbrace{3^3\!+(2\alpha^2)^3}^{\large 59}}{3\,+\,2\alpha^2\ \ \ }\, =\, 3^2-3(2\alpha^2)+(2\alpha^2)^2 =\, -6\alpha^2+8\alpha+9$$
A: Using the Euclidean algorithm we can show that
$$
\left(2x^2+3\right)\left(-\frac{6}{59}x^2+\frac{8}{59}x+\frac{9}{59}\right)-\left(x^3-2\right)\left(-\frac{12}{59}x+\frac{16}{59}\right)=1
$$
now replace $x=\alpha $ to get 
$$
\left(2\alpha ^2+3\right)\left(-\frac{6}{59}\alpha ^2+\frac{8}{59}\alpha +\frac{9}{59}\right)=1
$$
A: 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:
$$\frac{1}{3+2 \alpha^2} = \frac{(3+2 (\zeta_3 \alpha)^2) (3 + 2 (\zeta_3^2 \alpha)^2)}{(3+2 \alpha^2)(3+2 (\zeta_3 \alpha)^2) (3 + 2 (\zeta_3^2 \alpha)^2)}$$
$$=\frac{9+6 \zeta_3 \alpha^2+6 \zeta_3^2 \alpha^2 + 4 \zeta_3^3 \alpha^4}{(3+2 \alpha^2)(3+2 (\zeta_3 \alpha)^2) (3 + 2 (\zeta_3^2 \alpha)^2)}=\frac{9+8 \alpha - 6 \alpha^2}{(3+2 \alpha^2)(3+2 (\zeta_3 \alpha)^2) (3 + 2 (\zeta_3^2 \alpha)^2)}$$
$$=\frac{9+8 \alpha - 6 \alpha^2}{(3+2 \alpha^2)(9+8 \alpha - 6 \alpha^2)}=\frac{9+8 \alpha - 6 \alpha^2}{59}.$$
A: ${\rm mod}\ \alpha^3\!-\!2\!:\,\ 1\equiv (2\alpha^2\!+\!3)(a\alpha^2\!+\!b\alpha\!+\!c)\equiv (2\color{#c00}c\!+\!3a)\alpha^2\!+(3\color{#0a0}b\!+\!4a)\alpha\!+\!3c\!+\!4b$  
Therefore $\ \color{#c00}c \equiv -3a/2,\,\ \color{#0a0}b \equiv -4a/3\ $ so $\ 1 \equiv 3c\!+\!4b \equiv -59a/6,\ $ so $\ a \equiv -6/59$
