Finding the inverse of an element in $\mathbb{Q}(2^{\frac{1}{3}})$

Question

I did find the following post: Finding a representation of the inverse as linear combination

It is about the question how to calculate $(7+2^{\frac{2}{3}})^{-1} \in \mathbb{Q}(2^{\frac{1}{3}})$ as a $\mathbb{Q}$- linear combination of $\{1,2^{\frac{1}{3}},2^{\frac{2}{3}}\}$.

The below answer did suggest that one writes

$1=(7+2^{\frac{2}{3}})^{-1} (7+2^{\frac{2}{3}})=(\lambda_1+\lambda_22^{\frac{1}{3}}+\lambda_3 2^{\frac{2}{3}})(7+2^{\frac{2}{3}})$

I did try this approach but couldn't solve it correctly. Here is what I did:

$1=(7+2^{\frac{2}{3}})^{-1} (7+2^{\frac{2}{3}})=(\lambda_1+\lambda_2 2^{\frac{1}{3}}+\lambda_3 2^{\frac{2}{3}})(7+2^{\frac{2}{3}})$

$(\lambda_1+\lambda_22^{\frac{1}{3}}+\lambda_3 2^{\frac{2}{3}})(7+2^{\frac{2}{3}})=7\lambda_1+7 \lambda_2 2^{\frac{1}{3}}+7 \lambda_3 2^{\frac{2}{3}}+ \lambda_1 2^{\frac{2}{3}}+2 \lambda_2+2 \lambda_3 2^{\frac{1}{3}}$

$=(7\lambda_1+2\lambda_2)+2^{\frac{1}{3}}(\lambda_2+2\lambda_3)+2^{\frac{2}{3}}(7\lambda_3+\lambda_1)$

Thus we have $\;1=(7\lambda_1+2\lambda_2)+2^{\frac{1}{3}}(\lambda_2+2\lambda_3)+2^{\frac{2}{3}}(7\lambda_3+\lambda_1)$

and we get the system

$7\lambda_1+2\lambda_2=1$

$\lambda_2+2\lambda_3=0$

$\lambda_1+7\lambda_3=0$

Solving it (by hand and by Wolframalpha: Wolframalpha ) I get

$\lambda_1=\frac{7}{53}$

$\lambda_2=\frac{2}{53}$

$\lambda_3=-\frac{1}{53}$

Now checking the calculation I get that it is wrong: Wolframalpha )

Question: Where is my mistake? I don't really understand what went wrong. Please help me solve this.

Answer 1

Where is my mistake?

It should be $\color{red}7\lambda_2+2\lambda_3=0$, so $\lambda_1=\dfrac{49}{347}, \lambda_2=\dfrac2{347},$ and $\lambda_3=-\dfrac7{347}$.

Answer 2

Write out the product $$ \begin{align} 1 &=\left(7+2^{2/3}\right)\left(a+b2^{1/3}+c2^{2/3}\right)\\ &=(7a+2b)+(7b+2c)2^{1/3}+(a+7c)2^{2/3} \end{align} $$ which leads to the matrix equation $$ \begin{bmatrix}7&2&0\\0&7&2\\1&0&7\end{bmatrix}\begin{bmatrix}a\\b\\c\end{bmatrix}=\begin{bmatrix}1\\0\\0\end{bmatrix} $$ which gives the solution $$ \frac{49}{347}+\frac2{347}2^{1/3}-\frac7{347}2^{2/3} $$ The second equation (coefficient of $2^{1/3}$) seems to be where your mistake is