(Multiplicative) inverse of $\alpha = (\sqrt[3]{7})^2 + 3\sqrt[3]{7} + 1 \in\mathbb Q[\sqrt[3]{7}]$ Set $\mathbb Q[\sqrt[3]{7}] = \{F(\sqrt[3]{7}) \mid F ∈ Q[x]\}$ is a field (with the usual addition and the usual multiplication).
Calculate the (multiplicative) inverse of
$$\alpha = (\sqrt[3]{7})^2 + 3\sqrt[3]{7} + 1 \in \mathbb Q[\sqrt[3]{7}]$$
Note: Application of the Euclidean algorithm to the polynomials $x^2 + 3x + 1$ and $x^3 - 7$ could help.
Attempt:
I know there is a multiplcative inverse $\beta$ with $\alpha\beta=1$, and that one should exist in $\mathbb Q[\sqrt[3]{7}]$, but do not know how I would go about expressing it in any form simpler than $\frac1{(\sqrt[3]{7})^2 + 3\sqrt[3]{7} + 1}$.
How can I determine a simpler way to express the value of $\frac1{(\sqrt[3]{7})^2 + 3\sqrt[3]{7} + 1}$?

You actually have two possibilities:

*

*Either, denoting $x=\sqrt[3]7$, you try to find a linearcombination $ax^2+bx+c$ such that
$\;(ax^2+bx+)(x^2+3x+1)=1$, which leads to solving the linear system
$$\begin{cases}a+3b+c=0\\7a+b+3c=0\\21a+7b+c=1\end{cases},$$
which can be solved  finding the reduced row echelon form of the augmented matrix
$$\left[\begin{array}{rrr|l}1&3&1&0\\ 7&1&3&0\\ 21&7&1&1\end{array}\right].$$

*Or, extending a bit the hint, you apply the extended Euclidean algorithm to the polynomials $X^3-7$ and $X^2+3X+1$, which are coprime, to obtain a Bézout's relation
$$u(X)(X^2+3X+1)+v(X)(X^3-1)=1,$$
which, substituting $x$ to $X$, shows the inverse of $x^2+3x+1$ is $u(x)$.

