Find the inverse of $5+4\sqrt[3]{2}+3\sqrt[3]{4}$? 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 $(5a+6b+8c)+(4a+5b+6c)\sqrt[3]{2}+(3a+4b+5c)\sqrt[3]{4}=1$
 A: From your last line you get three simultaneous equations for $a,b,c$ which you solve
$$5a+6b+8c=1 \\4a+5b+6c=0 \\3a+4b+5c=0\\a+b+c=0\\b+3c=1\\b+2c=0\\c=1\\b=-2\\a=1$$
A: $$
\left(
\begin{array}{ccc|c}
5 & 6 & 8 & 1 \\
4 & 5 & 6 & 0 \\
3 & 4 & 5 & 0
\end{array}
\right)
$$
$$
\left(
\begin{array}{ccc|c}
1 & 1 & 2 & 1 \\
4 & 5 & 6 & 0 \\
3 & 4 & 5 & 0
\end{array}
\right)
$$
$$
\left(
\begin{array}{ccc|c}
1 & 1 & 2 & 1 \\
1 & 1 & 1 & 0 \\
3 & 4 & 5 & 0
\end{array}
\right)
$$
$$
\left(
\begin{array}{ccc|c}
0 & 0 & 1 & 1 \\
1 & 1 & 1 & 0 \\
3 & 4 & 5 & 0
\end{array}
\right)
$$
$$
\left(
\begin{array}{ccc|c}
0 & 0 & 1 & 1 \\
1 & 1 & 1 & 0 \\
0 & 1 & 2 & 0
\end{array}
\right)
$$
$$
\left(
\begin{array}{ccc|c}
0 & 0 & 1 & 1 \\
1 & 1 & 1 & 0 \\
0 & 1 & 0 & -2
\end{array}
\right)
$$
$$
\left(
\begin{array}{ccc|c}
0 & 0 & 1 & 1 \\
1 & 0 & 1 & 2 \\
0 & 1 & 0 & -2
\end{array}
\right)
$$
$$
\left(
\begin{array}{ccc|c}
0 & 0 & 1 & 1 \\
1 & 0 & 0 & 1 \\
0 & 1 & 0 & -2
\end{array}
\right)
$$
Apparently 
$$  a = 1, \; \; b = -2, \; \; c = 1 \; \; . $$
A: 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:
$$
-1 = \big(-5-4\sqrt[3]{2}-3\sqrt[3]{4}\big)\cdot\big(1 - 2\sqrt[3]{2}+\sqrt[3]{4}\big) \;\;\iff\;\; \frac{1}{5+4\sqrt[3]{2}+3\sqrt[3]{4}} = 1 - 2\sqrt[3]{2}+\sqrt[3]{4}
$$

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

$$
\frac{1}{x}=x^2 - 15 x + 3 = \big(5+4\sqrt[3]{2}+3\sqrt[3]{4}\big)^2 - 15 (5+4\sqrt[3]{2}+3\sqrt[3]{4}\big) + 3 = \ldots
$$
A: Mostly an observation:
$$ 5 + 4 \sqrt[3]{2}+3\sqrt[3]{4}=(1+\sqrt[3]{2}+\sqrt[3]{4})^2$$
so 
$$(5 + 4 \sqrt[3]{2}+3\sqrt[3]{4})^{-1}=((1+\sqrt[3]{2}+\sqrt[3]{4})^{-1})^2=(\sqrt[3]{2}-1)^2= 1- 2\sqrt[3]{2}+\sqrt[3]{4}$$
