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.
The Paper in question: http://www.ams.org/journals/bull/2004-41-01/S0273-0979-03-00993-5/S0273-0979-03-00993-5.pdf 
Relevant section is towards the end of second page.
 A: 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). 
