 Solution

Let $\alpha =\sqrt[3]{2}$. Its minimal polynomial is $f(X)=X^3-2$, whose discriminant is $-108$. Since $5\nmid 108$, the prime $5$ does not divide the index $[{\mathcal{O}}_K:\mathbb{Z}[\alpha ]]$, so the Dedekind factorization theorem applies at $5$. In ${\mathbb{F}}_5[X]$,The quadratic factor has discriminant $3$, which is a quadratic nonresidue modulo $5$, so it is irreducible. ThereforewhereTheir residue-field degrees are one and two. Both factors of $f$ occur with multiplicity one, so both prime-ideal exponents are one. Hence neither ${\mathfrak{p}}_1$ nor ${\mathfrak{p}}_2$ is ramified.