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

$$
\left(
\begin{array}{ccc|c}
5 & 6 & 8 & 1 \\
4 & 5 & 6 & 0 \\
3 & 4 & 5 & 0
\end{array}
\right)
$$
$$
\left(
\begin{array}{ccc|c}
1 & 1 & 2 & 1 \\
4 & 5 & 6 & 0 \\
3 & 4 & 5 & 0
\end{array}
\right)
$$
$$
\left(
\begin{array}{ccc|c}
1 & 1 & 2 & 1 \\
1 & 1 & 1 & 0 \\
3 & 4 & 5 & 0
\end{array}
\right)
$$
$$
\left(
\begin{array}{ccc|c}
0 & 0 & 1 & 1 \\
1 & 1 & 1 & 0 \\
3 & 4 & 5 & 0
\end{array}
\right)
$$
$$
\left(
\begin{array}{ccc|c}
0 & 0 & 1 & 1 \\
1 & 1 & 1 & 0 \\
0 & 1 & 2 & 0
\end{array}
\right)
$$
$$
\left(
\begin{array}{ccc|c}
0 & 0 & 1 & 1 \\
1 & 1 & 1 & 0 \\
0 & 1 & 0 & -2
\end{array}
\right)
$$
$$
\left(
\begin{array}{ccc|c}
0 & 0 & 1 & 1 \\
1 & 0 & 1 & 2 \\
0 & 1 & 0 & -2
\end{array}
\right)
$$
$$
\left(
\begin{array}{ccc|c}
0 & 0 & 1 & 1 \\
1 & 0 & 0 & 1 \\
0 & 1 & 0 & -2
\end{array}
\right)
$$
Apparently 
$$  a = 1, \; \; b = -2, \; \; c = 1 \; \; . $$
