# (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 $$\frac1{(\sqrt[3]{7})^2 + 3\sqrt[3]{7} + 1}$$?

• Use the hint. It will work. Dec 14, 2020 at 12:23
• Welcome! Please use MathJax (Latex syntax) to make your question more readable! Have you tried the hint? As Wuestenfux remarked: Those tend to help
– CPCH
Dec 14, 2020 at 13:00
• $$\frac{1}{1+3 \cdot\sqrt[3]{7}+7^{2/3}}=\frac{1}{44} \left(-5+\sqrt[3]{7}+2\cdot 7^{2/3}\right)$$ Dec 14, 2020 at 15:34

• 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 $$u(x)$$.
Alternatively, $$\alpha$$ is a root of $$x^3 - 3 x^2 - 60 x - 176$$ and so its inverse is $$\frac{1}{176}(\alpha^2-3\alpha-60)$$.