In general, if the ring of integers of $\mathcal{O}_K$ is of the form $\mathbb{Z}[\alpha]$, and $q(x)$ is the irreducible polynomial of $\alpha$ over $\mathbb{Q}$, then there is a natural bijection between the primes in $\mathcal{O}_K$ that lie above the rational prime $p$ and the irreducible factors of $\overline{q}(x)$ in $\mathbb{F}_p$, where $\overline{q}$ is the image of $q(x)$ under reduction modulo $p$. The bijection is that a prime $\mathfrak{p}$ of $\mathcal{O}_K$ lies above $(p)$ corresponds to the factor $\overline{Q}(x)$ if and only if $\mathfrak{p}$ is the kernel of the map $\mathbb{Z}[\alpha]\to\mathbb{F}_p[\alpha]$. So the factorization of $(p)$ in $\mathcal{O}_K$ exactly mimicks the factorization of $\overline{q}(x)$ over $\mathbb{F}_p$: same number of factors, same repetitions, etc.
So you want to factor $x^3-2$ over $\mathbb{F}_p$; this suffices here because if $\alpha$ is a root of $x^3-2$, then $\mathcal{O}_K = \mathbb{Z}[\alpha]$ (see for example Problem 41(a)-(d) in Chapter 2 of Marcus's Number Fields). Edit: I was working in $\mathbb{Q}(\sqrt[3]{2})$ instead of the splitting field; sorry. But from the fact that they ramify in this subextension, you know they will ramify in $K$; the only question is whether they ramify as $\mathfrak{p}^6$ or as $\mathfrak{p_1}^3\mathfrak{p_2}^3$, depending on whether the prime above $p$ splits or ramifies, or is inert in $K$. Since you can go from $\mathbb{Q}(\sqrt[3]{2})$ to $K$ by adjoining $\omega$, and the ring of integers will be, if I'm not mistaken, $\mathbb{Z}[\alpha][i]$, you can use the same theorem above (which holds for arbitrary integral extensions, not just over $\mathbb{Z}$) to get that next step, looking at $x^2+x+1$.
With $p=2$, the polynomial factors over $\mathbb{F}_p$ as $x^3$, so $(2) = \mathfrak{p}^3$ for some prime $\mathfrak{p}$ of $\mathcal{O}_K$; for $p=3$, you have $x^3 + 2 = x^3 - 1 = (x-1)^3$, so again $(3)=\mathfrak{q}^3$ for some prime $\mathfrak{q}$ of $\mathcal{O}_K$.