The problem above has many answers in StackExchange. I'm trying to understand this specific one.
He mentions the formula $\text{disc}(\mathbb{Z}[\alpha])=(\mathcal{O}:\mathbb{Z}[\alpha])^2\text{disc}(\mathcal{O})$, computes $\text{disc}(\mathbb{Z}[\alpha])=-2^23^3$, then concludes that $6\mathcal{O}\subset \mathbb{Z}[\alpha]$ (here $\alpha:=\sqrt[3]{2}$).
The only part I don't understand why is "$6\mathcal{O}\subset \mathbb{Z}[\alpha]$"
Any hints?
From the mentioned formula, you get $(\mathcal O:\mathbb Z[\alpha])^2\mid\mathrm{disc}(\mathbb Z[\alpha])=-2^23^3$. What can you conclude about $(\mathcal O:\mathbb Z[\alpha])$?
From the identity $$(\mathcal{O}:\Bbb{Z}[\alpha])^2\cdot\operatorname{disc}(\mathcal{O}) =\operatorname{disc}(\Bbb{Z}[\alpha]) =-2^23^3,$$ it follows that $(\mathcal{O}:\Bbb{Z}[\alpha])$ divides $6$. That means precisely that the order of the quotient of abelian groups $\mathcal{O}/\Bbb{Z}[\alpha]$ divides $6$. In particular $6\equiv0$ in the quotient $\mathcal{O}/\Bbb{Z}[\alpha]$ and so $6x\in\Bbb{Z}[\alpha]$ for all $x\in\mathcal{O}$, meaning that $6\mathcal{O}\subset\Bbb{Z}[\alpha]$.
