= Solution
Let $\alpha=\sqrt[3]{2}$. Its minimal polynomial is $f(X)=X^3-2$, whose discriminant is $-108$. Since $5\nmid108$, the prime $5$ does not divide the index $[\mathcal O_K:\mathbb Z[\alpha]]$, so the <Dedekind factorization theorem> applies at $5$. In $\mathbb F_5[X]$,
$$
X^3-2=(X-3)(X^2+3X+4).
$$
The quadratic factor has discriminant $3$, which is a <quadratic nonresidue> modulo $5$, so it is irreducible. Therefore
$$
(5)=\mathfrak p_1\mathfrak p_2,
$$
where
$$
\mathfrak p_1=(5,\alpha-3),
\qquad
\mathfrak p_2=(5,\alpha^2+3\alpha+4).
$$
Their 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.
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