Find the inverse of $5+4\sqrt[3]{2}+3\sqrt[3]{4}$?

Question

I'm trying to find the inverse of $5+4\sqrt[3]{2}+3\sqrt[3]{4}$ in $\mathbb{Z}[\sqrt[3]{2}]$. I know it is a unit, so there is an inverse, but I feel like I may be doing too much work in the wrong direction. Here's what I have so far:

Let $\alpha = 5+4\sqrt[3]{2}+3\sqrt[3]{4}$ and $\alpha^{-1} = a+b\sqrt[3]{2}+c\sqrt[3]{4}$ for some $a,b,c \in \mathbb{Z}$.

$(a+b\sqrt[3]{2}+c\sqrt[3]{4})(5+4\sqrt[3]{2}+3\sqrt[3]{4})=1$

$=5a+6b+8c+4a\sqrt[3]{2}+5b\sqrt[3]{2}+6c\sqrt[3]{2}+3a\sqrt[3]{4}+4b\sqrt[3]{4}+5c\sqrt[3]{4}=1$

$=a(5+4\sqrt[3]{2}+3\sqrt[3]{4})+b(6+5\sqrt[3]{2}+4\sqrt[3]{4})+c(8+6\sqrt[3]{2}+5\sqrt[3]{4})=1$

and trying to solve for a,b, and c, but I don't know how?

Edit: Regrouping to $(5a+6b+8c)+(4a+5b+6c)\sqrt[3]{2}+(3a+4b+5c)\sqrt[3]{4}=1$

Answer 1

From your last line you get three simultaneous equations for $a,b,c$ which you solve $$5a+6b+8c=1 \\4a+5b+6c=0 \\3a+4b+5c=0\\a+b+c=0\\b+3c=1\\b+2c=0\\c=1\\b=-2\\a=1$$

Answer 2

$$ \left( \begin{array}{ccc|c} 5 & 6 & 8 & 1 \\ 4 & 5 & 6 & 0 \\ 3 & 4 & 5 & 0 \end{array} \right) $$ $$ \left( \begin{array}{ccc|c} 1 & 1 & 2 & 1 \\ 4 & 5 & 6 & 0 \\ 3 & 4 & 5 & 0 \end{array} \right) $$ $$ \left( \begin{array}{ccc|c} 1 & 1 & 2 & 1 \\ 1 & 1 & 1 & 0 \\ 3 & 4 & 5 & 0 \end{array} \right) $$ $$ \left( \begin{array}{ccc|c} 0 & 0 & 1 & 1 \\ 1 & 1 & 1 & 0 \\ 3 & 4 & 5 & 0 \end{array} \right) $$ $$ \left( \begin{array}{ccc|c} 0 & 0 & 1 & 1 \\ 1 & 1 & 1 & 0 \\ 0 & 1 & 2 & 0 \end{array} \right) $$ $$ \left( \begin{array}{ccc|c} 0 & 0 & 1 & 1 \\ 1 & 1 & 1 & 0 \\ 0 & 1 & 0 & -2 \end{array} \right) $$ $$ \left( \begin{array}{ccc|c} 0 & 0 & 1 & 1 \\ 1 & 0 & 1 & 2 \\ 0 & 1 & 0 & -2 \end{array} \right) $$ $$ \left( \begin{array}{ccc|c} 0 & 0 & 1 & 1 \\ 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & -2 \end{array} \right) $$

Apparently $$ a = 1, \; \; b = -2, \; \; c = 1 \; \; . $$

Answer 3

Alt. hint:   let $\,x=5+4\sqrt[3]{2}+3\sqrt[3]{4}\,$, then successively multiplying with $\,\sqrt[3]{2}\,$:

$$ \begin{cases} \begin{align} 5-x+4\sqrt[3]{2}+3\sqrt[3]{4} &= 0 \\ 12+(5-x)\sqrt[3]{2}+4 \sqrt[3]{4} &= 0 \\ 8 + 12 \sqrt[3]{2}+(5-x)\sqrt[3]{4} &= 0 \end{align} \end{cases} $$

Eliminating $\,\sqrt[3]{2}\,$ and $\,\sqrt[3]{4}\,$ between the equations gives in the end:

$$ x^3 - 15 x^2 + 3 x - 1 = 0 \tag{1} $$

Dividing the polynomial by the known factor $\,x-5-4\sqrt[3]{2}-3\sqrt[3]{4}\,$ gives the factorization:

$$ x^3 - 15 x^2 + 3 x - 1 \\ = \big(x-5-4\sqrt[3]{2}-3\sqrt[3]{4}\big)\cdot\big(x^2 + (-10 + 4 \sqrt[3]{2} + 3 \sqrt[3]{4}) x + 1 - 2\sqrt[3]{2}+\sqrt[3]{4}\big) $$

Finally, identifying the free terms between the two sides gives:

$$ -1 = \big(-5-4\sqrt[3]{2}-3\sqrt[3]{4}\big)\cdot\big(1 - 2\sqrt[3]{2}+\sqrt[3]{4}\big) \;\;\iff\;\; \frac{1}{5+4\sqrt[3]{2}+3\sqrt[3]{4}} = 1 - 2\sqrt[3]{2}+\sqrt[3]{4} $$

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[ EDIT ]   Another way to conclude, while avoiding the polynomial division, is to write $(1)$ as:

$$ \frac{1}{x}=x^2 - 15 x + 3 = \big(5+4\sqrt[3]{2}+3\sqrt[3]{4}\big)^2 - 15 (5+4\sqrt[3]{2}+3\sqrt[3]{4}\big) + 3 = \ldots $$

Answer 4

Mostly an observation: $$ 5 + 4 \sqrt[3]{2}+3\sqrt[3]{4}=(1+\sqrt[3]{2}+\sqrt[3]{4})^2$$ so $$(5 + 4 \sqrt[3]{2}+3\sqrt[3]{4})^{-1}=((1+\sqrt[3]{2}+\sqrt[3]{4})^{-1})^2=(\sqrt[3]{2}-1)^2= 1- 2\sqrt[3]{2}+\sqrt[3]{4}$$

Answer 5

Let $\alpha=5+4\sqrt[3]{2}+3\sqrt[3]{4}$. Then \begin{align} \alpha\sqrt[3]{2}&=6+5\sqrt[3]{2}+4\sqrt[3]{4}\\ \alpha\sqrt[3]{4}&=8+6\sqrt[3]{2}+5\sqrt[3]{4} \end{align} Thus you need to find the inverse matrix of \begin{bmatrix} 5 & 6 & 8 \\ 4 & 5 & 6 \\ 3 & 4 & 5 \end{bmatrix} With Gaussian elimination \begin{align} \begin{bmatrix} 5 & 6 & 8 & 1 & 0 & 0 \\ 4 & 5 & 6 & 0 & 1 & 0 \\ 3 & 4 & 5 & 0 & 0 & 1 \end{bmatrix} &\to \begin{bmatrix} 1 & 6/5 & 8/5 & 1/5 & 0 & 0 \\ 0 & 1/5 & -2/5 & -4/5 & 1 & 0 \\ 0 & 2/5 & 1/5 & -3/5 & 0 & 1 \end{bmatrix} \\&\to \begin{bmatrix} 1 & 6/5 & 8/5 & 1/5 & 0 & 0 \\ 0 & 1 & -2 & -4 & 5 & 0 \\ 0 & 2 & 1 & -3 & 0 & 5 \end{bmatrix} \\&\to \begin{bmatrix} 1 & 6/5 & 8/5 & 1/5 & 0 & 0 \\ 0 & 1 & -2 & -4 & 5 & 0 \\ 0 & 0 & 5 & 5 & -10 & 5 \end{bmatrix} \\&\to \begin{bmatrix} 1 & 6/5 & 8/5 & 1/5 & 0 & 0 \\ 0 & 1 & -2 & -4 & 5 & 0 \\ 0 & 0 & 1 & 1 & -2 & 1 \end{bmatrix} \\&\to \begin{bmatrix} 1 & 6/5 & 0 & -7/5 & 16/5 & -8/5 \\ 0 & 1 & 0 & -2 & 1 & 2 \\ 0 & 0 & 1 & 1 & -2 & 1 \end{bmatrix} \\&\to \begin{bmatrix} 1 & 0 & 0 & 1 & 2 & -4 \\ 0 & 1 & 0 & -2 & 1 & 2 \\ 0 & 0 & 1 & 1 & -2 & 1 \end{bmatrix} \end{align} we see that the inverse is \begin{bmatrix} 1 & 2 & -4 \\ -2 & 1 & 2 \\ 1 & -2 & 1 \end{bmatrix} and the inverse is $$ 1-2\sqrt[3]{2}+\sqrt[3]{4} $$ corresponding to the first column, because this gives the effect of multiplying by the inverse of $\alpha$.

Explanation: the matrix I wrote at the beginning is the matrix of the linear map $\mathbb{Q}[\sqrt[3]{2}]$ induced by the multiplication by $\alpha$. Its inverse is the matrix of the linear map induced by the multiplication by $\alpha^{-1}$.