Finding an integral basis for a cubic field using the discriminant

Question

This is sort of a revival of this dead post but with a few tangential questions that I don't feel like I should coattail onto the linked post. Suppose we are trying to find an integral basis for $\mathbb{Q}[\alpha]$ where $\alpha^{3}-2 = 0$. We find that $d = \text{disc}(\lbrace 1, \alpha, \alpha^{2} \rbrace) = -108 = -2^{2}3^{3}$ and since this isn't squarefree, we have more work to do if we're to show $\lbrace 1, \alpha, \alpha^{2} \rbrace$ is an integral basis. So, we can use the result that $$ \mathbb{Z}[\alpha] \subset \mathcal{O}_{\mathbb{Q}[\alpha]} \subset d^{-1}\mathbb{Z}[\alpha]. $$ Now, what I'd like to do is to show that for any $\beta \in d^{-1}\mathbb{Z}[\alpha] \setminus \mathbb{Z}[\alpha]$ we have $\beta$ is not integral. As per the linked post, we should only verify that $\frac{1+\alpha+\alpha^{2}}{x}$ is not integral for $x=2,3$. First question: why are these the only elements that we need to check? It seems to me like we should check this for $\frac{a+b\alpha+c\alpha^{2}}{x}$ for $x | 2^{2}3^{3}$ and $a,b,c \in \mathbb{Z}$. On top of that, is there a clever way to compute the norm of a generic element of $d^{-1}\mathbb{Z}[\alpha]$ that I'm missing, because I don't know how I'd go about this.