The norm $N:\mathbb{Z}[\sqrt[3]{2}] \rightarrow \mathbb{N}$ defined by $N(a+b\sqrt[3]{2} + c\sqrt[3]{4}) = |a^3 + 2b^3 + 4c^3 - 6abc|$ is multiplicative. (Already proven). Show that $\alpha \in \mathbb{Z}[\sqrt[3]{2}]$ is a unit if and only if $N(\alpha) = 1$.
So far proved it one direction, but am struggling with the other.
Proof $\rightarrow$ Let $\alpha$ be a unit, so by defintion $\exists \beta$ such that $\alpha\beta = 1$. We can then use the norm, knowing that $N(\alpha\beta) = |(1)^3 + 2(0)^3 + 4(0)^3 - 6(1)(0)(0)| = 1$. Since the norm is multiplicative, $N(\alpha\beta) = N(\alpha)N(\beta) = 1$, and since the results of the norms have to be positive integers, we know $N(\alpha)=1$.
Proof $\leftarrow$ Let $N(\alpha)$ = 1. Then for $\alpha = a+b\sqrt[3]{2} + c\sqrt[3]{4}$, $|a^3 + 2b^3 + 4c^3 - 6abc|=1$
Any help on where to go?
The fact that $\;N(\alpha)=\pm1\;$ means that the minimal polynomial of $\;\alpha\;$ over the rationals is of the form
$$p(x)=x^n+a_{n-1}x^{n-1}+\ldots+a_1x\pm1\;,\;\;a_k\in\Bbb Z$$
Taking the reciprocal of the above polynomial we get a polynomial that vanishes at $\;\frac1\alpha\;$ (this much is always true whenever $\;\alpha\neq0\;$ and we're working over a field), thus $\;\frac1\alpha\;$ is a root of
$$f(x)=\pm x^n+a_1x^{n-1}+\ldots+a_{n-1}x+1$$
and thus $\;\frac1\alpha\;$ is an algebraic integer.