Let $K = \mathbb{Q}(\sqrt[3]{6})$. Factorise $\langle p \rangle$ into prime ideals in $\mathcal{O}_K$ for $p = 2, 5, 13$, checking that the factors are principal.
I used the Dedekind-Kummer Theorem to get the factors, but I haven't been able to show that all of them are principal. I've managed to do the factors of $\langle 2 \rangle, \langle 5 \rangle$ using some difference of cubes stuff, but I'm not sure how one would find generators for the factors of $\langle 13 \rangle$. For example, $\langle 13, -2+\sqrt[3]{6} \rangle$ (the others have $-7,-12$ instead of $-2$; I hope these are correct). I suspect it's something relatively simple but I can't see it as none of the obvious ones like $-2+\sqrt[3]{6}$ or $11+\sqrt[3]{6}$ seem to work. Help would be appreciated.
The ideal $\langle13\rangle\subset\mathcal{O}_K$ is itself prime because $X^3-6$ is irreducible over $\Bbb{F}_{13}$, which is easily seen by checking that it has no roots in $\Bbb{F}_{13}$. In particular, to answer your question, the unique factor $\langle13\rangle$ is principal.