Prove $u$ is a unit if and only if $N(u) = 1$? 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?
 A: $R=\Bbb Z[\sqrt[3]2]$ is a free Abelian group of rank $3$, that is isomorphic
as an Abelian group to $\Bbb Z^3$. A convenient basis for this is
$1$, $\sqrt[3]2$ and $\sqrt[3]4$.
If $\alpha\in R$, multiplication by $\alpha$ determines an endomorphism
of $R$; a group homomorphism from $R$ to $R$. We can represent this
by a matrix by considering its action on one's favourite basis
(say, $1$, $\sqrt[3]2$, $\sqrt[3]4$). Then $N(\alpha)=|\det A|$.
(You can prove this by the explicit formula you already have).
If $N(\alpha)=1$, then $\det A=\pm1$. An endomorphism of $\Bbb Z^3$
with determinant $1$ is an automorphism, so a bijection. In terms of
$A$ then multiplication by $\alpha$ is a bijection. Therefore, there is
$\beta\in R$ with $\alpha\beta=1$.
A: Hint: if you embed $\mathbb{Z}[\sqrt[3]{2}]$ into $\mathbb{Z}[\sqrt[3]{2}, \omega]$ where $\omega := e^{2 \pi i / 3}$, then
$$N(a + b \sqrt[3]{2} + c \sqrt[3]{4}) = \left| (a + b \sqrt[3]{2} + c \sqrt[3]{4}) (a + b \omega \sqrt[3]{2} + c \omega^2 \sqrt[3]{4}) (a + b \omega^2 \sqrt[3]{2} + c \omega \sqrt[3]{4}) \right|.$$
(This expression is heavily inspired by the definition of norm in Galois theory, of which your definition is close to being a special case.)
Now, multiply the last two factors on the right hand side, and you should get an expression for $\pm$ of the inverse of $a + b \sqrt[3]{2} + c \sqrt[3]{4}$ which lies in $\mathbb{Z}[\sqrt[3]{2}]$.  (With the sign being determined by the sign of $a^3 + 2 b^3 + 4 c^3 - 6abc$.)
A: 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.
A: As mentioned by DonAntonio, the minimal polynomial of $\alpha$ is of the form
$$p(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_1x\pm1,\quad a_k\in\Bbb Z$$
Then $p(\alpha)=0$ implies
$$
 \alpha (\alpha^{n-1}+\cdots+a_2\alpha+a_1) = \pm 1
$$
and so the inverse of $\alpha$ is $\alpha^{n-1}+\cdots+a_2\alpha+a_1 \in \mathbb Z[\alpha] \subseteq \mathbb{Z}[\sqrt[3]{2}]$.
A: For $K$ a number field the norm $N(\alpha)$ of an element $\alpha$ of $K$ is the determinant of the $\mathbb{Q}$-linear map $m_{\alpha} :x \mapsto \alpha x$. Assume that $\alpha$ is inside the ring of integers $\mathcal{O}$ of $K$. In a $\mathbb{Z}$ basis of $\mathcal{O}$ the map $m_{\alpha}$ has a matrix with integral coefficients. Therefore $\alpha$ satisfies an equation of the form
$$\alpha^n - a_1 \alpha^{n-1} + \cdots + (-1)^n a_n= 0$$ with $a_i \in\mathbb{Z}$. Note that $a_n= N(\alpha)$. Assume now that $N(\alpha)=\pm 1$. Then we can write 
$$\alpha( \alpha^{n-1} - a_1 \alpha^{n-2} + \cdots) = \pm 1$$ Therefore the inverse $\alpha^{-1}$  is in $\mathbb{Z}[\alpha]$ and so an integer.
