Let $\alpha=\sqrt[3]{19}$, $K=\mathbb Q(\alpha)$ and ring of integer is given $\mathcal O_K$.
Now I want to calculate some prime decomposition. However, in Marcus Number Fields' notes, since extension degree $3$ we cannot do the factoring $$x^3-19\equiv \cdots \mod 3$$
For the ideal $(3)\mathcal O_K$
My idea is, if possible, to find another minimum polynomial which gives the root $\alpha=\sqrt[3]{19}$ but also gives the other conjugates such that $\mathcal O_K$ and $\mathbb Z[\sqrt[3]{19}]$ meets.
So how to construct such polynomial?
I tried to factor with silly fashion, using integral basis of $\mathcal O_K$ of $\mathbb Q(\sqrt[3]{19})$ $$\mathcal O_K=\mathbb Z\left[1,\alpha,\frac{\alpha^2+\alpha+1}{3} \right]$$
as $$\left(x-1 \right)\left(x-\alpha \right)\left( x-\frac{\alpha^2+\alpha+1}{3}\right)\not\in \mathbb Z[x]$$
So how to calculate decomposition of $(3)$ in this cubic field?
Following your suggestion, the real root $\beta$ of $x^3-x^2-6x-12$ gives $K=\mathbb{Q}(\sqrt[3]{19})$. The absolute value of the discriminant of the basis generated by $\beta$ is $4332$, thus $[\mathcal{O}_K:\mathbb{Z}(\beta)]$ is 2 (because the actual discriminant is $4332/4$).
Since $3\nmid{[\mathcal{O}_K:\mathbb{Z}(\beta)]}$ we can apply the Dedekind-Kummer theorem:
$$x^3-x^2-6x-12\equiv{x^2(x-1)}\mod3.$$
Thus, $(3)=(3,\beta)^2(3, \beta-1)$.