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.

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

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

2 Answers 2


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 $u(x)$.

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


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