Show that no prime number $p$ stays prime in $\mathbb{Q}(\omega, \sqrt[3]{2})$, where $\omega^3 = 1$, $\omega \neq 1$.
My idea: Instead of trying to compute the ring of integers of $\mathbb{Q}(\omega, \sqrt[3]{2})$, I computed the ring of integers of $\mathbb{Q}(\sqrt[3]{2})$: $\mathbb{Z}[\sqrt[3]{2}]$. Now I can analyze the factorization of primes in $\mathbb{Z}[\sqrt[3]{2}]$. In order to do so, I can look at how $x^3 - 2$ factorize $\pmod p$. However, I am having troubles to complete this last step.
Here’s another argument, (slightly) more elementary than that given by @reuns in their comment:
Let $K\supset k$ be a Galois extension of number fields, say with rings of integers $\mathcal O\supset \mathfrak o$. Suppose also that $\mathfrak p$ is a prime downstairs that is unramified and not split that is, that $\mathfrak p\mathcal O$ is still a prime ideal of $\mathcal O$. Then $\text{Gal}^K_k$ is isomorphic to the corresponding Galois group in finite characteristic. But the latter, as Galois group of an extension of finite fields, is cyclic.
Thus if the prime downstairs remains prime upstairs, the Galois group has to be cyclic. But the Galois group of $\Bbb Q(\omega,\sqrt[3]2)$ over $\Bbb Q$ is $S_3$, not cyclic. So no prime downstairs remains prime upstairs.