Given a field $K$, for which separable $\alpha,\beta$ do we have $[K(\alpha,\beta):K]=[K(\alpha):K]\cdot [K(\beta):K]$? Given a field $K$, for which separable $\alpha,\beta$ do we have $[K(\alpha,\beta):K]=[K(\alpha):K]\cdot [K(\beta):K]$?
I encountered this question in multiple forms wile solving some algebra questions, but wasn't able to find a general answer. I do know that in general
$$[K(\alpha,\beta):K]\le[K(\alpha):K]\cdot [K(\beta):K]$$
and that equality holds if $[K(\alpha):K]$ and $[K(\beta):K]$ are coprime, but that's about it.

The criterion is linear disjointness, which you can look up.
The definition is: Two extensions $K_1$ and$K_2$ of $k$ are linearly disjoint over $k$ if a $k$-basis of $K_1$ remains linearly independent over $K_2$.
As an example of fields that are linearly disjoint over $\Bbb Q$, take $K_1=\Bbb Q(\sqrt2\,)$ and $K_2=\Bbb Q(i)$.
As an example of fields that are not linearly disjoint over $\Bbb Q$, take for $K_1$ the real field $\Bbb Q(\sqrt[3]2\,)$, and for $K_2$ the isomorphic but nonreal field $\Bbb Q(\omega\sqrt[3]2\,)$, where $\omega$ is a primitive cube root of unity, $\omega^2+\omega+1=0$.
You can check that linear disjointness of your two simple extensions of $K$ is exactly what’s needed for your degree formula to hold.
