The ring of integers of a cubic number field generated by a root of $x^3-2x-2$ 
Let $\alpha$ be a root of the polynomial $f(x) = x^3-2x-2$, and let $K = \mathbb{Q}(\alpha)$. Show that $\mathcal{O}_{K} = \mathbb{Z}[\alpha]$.

I know the following
$$\text{disc}(1,\alpha,\alpha^2)=-76$$
$$\text{disc}(1,\alpha,\alpha^2)=[\mathcal{O}_K:\mathbb{Z}[\alpha]]^2\text{disc}(K)$$
It follows that $[\mathcal{O}_K:\mathbb{Z}[\alpha]]$ equals $1$ or $2$. I don't know how to complete from there, any one can help!
Thanks in advance.
 A: There is a tedious way and a slick way.
Tedious way:  Since the index is $1$ or $2$, the quotient $\mathcal O_K/\mathbf Z[\alpha]$ has size $1$ or $2$.  Therefore the group is killed by doubling, so $2\mathcal O_K \subset \mathbf Z[\alpha]$.  Thus $\mathbf Z[\alpha] \subset \mathcal O_K \subset (1/2)\mathbf Z[\alpha]$, which means you can just run through coset representatives $\beta_i$ for $((1/2)\mathbf Z[\alpha])/\mathbf Z[\alpha]$ to see which ones are algebraic integers: if $\beta_i$ is an algebraic integer, then the whole coset $\beta_i + \mathbf Z[\alpha]$ consists of algebraic integers, and if $\beta_i$ isn't an algebraic integer, then nothing in $\beta_i + \mathbf Z[\alpha]$ is an algebraic integer.
As representatives $\beta_i$ use
$(a/2) + (b/2)\alpha + (c/2)\alpha^2$ where $a, b, c \in \{0,1\}$. That is $8$ numbers to check. Well, really it's $7$ numbers to check since there the representative $0$, corresponding to the coset $\mathbf Z[\alpha]$, obviously is inside $\mathcal O_K$.  Not a single one of those $7$ nonzero numbers turns out to be an algebraic integer. For example, $\alpha/2 + \alpha^2/2$ is a root of $x^3 - 2x^2 - x - 1/4$. (This must be the minimal polynomial of $\alpha/2 + \alpha^2/2$ over $\mathbf Q$, granting that it really has that number as a root, since every number outside of $\mathbf Q$ in a cubic field has degree $3$ over $\mathbf Q$.) The only coset in $((1/2)\mathbf Z[\alpha])/\mathbf Z[\alpha]$ consisting of algebraic integers is the zero coset, and that means $\mathcal O_K = \mathbf Z[\alpha]$.
Slick way: the number $\alpha$ is a root of an Eisenstein polynomial at $2$, so $[\mathcal O_K:\mathbf Z[\alpha]]$ is not divisible by $2$ by Theorem 2.3 here. Therefore $[\mathcal O_K:\mathbf Z[\alpha]]$ must be $1$, which means $\mathcal O_K = \mathbf Z[\alpha]$.
