Finding the units in $\mathbb{Z}[\sqrt[3]{2}]$ and other questions

I'm reading a paper in which they solve the equation: $$a^3-2b^3=\pm 1$$ in integers using algebraic number theory.

The number $a-b\alpha$, with $\alpha=\sqrt[3]{2}$, is a unit in $\mathbb{Z}[\alpha]$. The units of this ring are, up to sign, powers of the single unit $1+\alpha+\alpha^2$. With some work one finds that $|a-b\alpha|$ can only be the zero'th power, so that $a=\pm 1$ and $b=0$.

1) Why are the units of the ring only powers of $1+\alpha+\alpha^2$? I can't find anything to this effect in my textbook and web searches won't turn up with anything.

2) Can't $|a-b\alpha|$ be the $(-1)$th power as well and $a=b=\pm 1$?

The second question I've concluded is a minor error but I can't be satisfied with this solution without a proof for my first question.

Relevant section is towards the end of second page.

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I think they might be using Dirchlet's unit theorem here, and also showing that $1 + \alpha + \alpha^2$ is the fundamental unit. Check Dirichlet's unit theorem on the web. – Rankeya Nov 28 '12 at 2:51
Dirichlet's unit theorem definitely implies the group of units is isomorphic to $\mathbb{Z}$ because $\mathbb{Q}(2^{1/3})$ has one real embedding and 2 complex embeddings into $\mathbb{C}$. – Rankeya Nov 28 '12 at 2:53
Thanks for giving me something to go with. I'm about to look up the theorem now. – genepeer Nov 28 '12 at 3:10
Dirichlet's unit theorem is one of the most beautiful theorems in algebraic number theory. It lets you describe the structure of the group of units of nice rings in terms of geometry. – fretty Nov 28 '12 at 10:46

As noted in comments, Dirichlet's unit theorem shows that the group of units in $\mathbb Z[2^{1/3}]$ (which is the full ring of integers in $\mathbb Q(2^{1/3})$ --- see this question) consists of elements of the form $\pm \eta^n$ for some fundamental unit $\eta$. The proof of Dirichlet's theorem should be effective in principal (perhaps this book adopts a perspective which helps with making things effective), and I guess in practice for sufficiently simple fields (such as $\mathbb Q(2^{1/3})$). In any case, you can certainly use a computer algebra package (such as sage) to determine that $1+\alpha + \alpha^2$ is a fundamental unit (assuming that it is; I didn't check).
You are correct that $(1+\alpha + \alpha^2)^{-1} = -1 + \alpha,$ and so the authors of the paper you are reading mistated their claim. [In the context of the paper you are reading, note that $a = 1, b = 0$ actually leads to the solution $x = 0, y = 1$, and it is the omitted solution $a = b = 1$ which leads to $x = 2, y = 3$.]
Also, while it is easy to see that any power of $1+\alpha+\alpha^2$ has a non-zero coefficient of $\alpha^2$ (if we write $(1+\alpha+ \alpha^2)^n = a_n + b_n\alpha + c_n \alpha^2$ then there is a simple recursion for $a_n,b_n,c_n$ in terms of $a_{n-1}, b_{n-1},c_{n-1}$, and one sees that $a_n,b_n,c_n$ are always positive, because this recursion only involves addition, no subtraction, and $a_1 = b_1 = c_1 = 1$ is positive), this is less obvious (at least to me) for the negative powers, because while there is also a simple recursion for the coefficients of $(1-\alpha)^n$, in this case the recursion has a mixture of signs, and the coefficients do vary in sign, so you will have to work harder to verify that the coefficient of $\alpha^2$ is never zero when $n > 1$ (again, assuming that it's in fact true, which I didn't try to check).