After starting to study field extensions I've had to do many exercises where one has to find the degree and a basis of a given field extension. I'll take $\mathbb{Q}(\sqrt{2} + \sqrt[3]{4})$ as an example. First one could see that $\mathbb{Q}(\sqrt{2} + \sqrt[3]{4}) = \mathbb{Q}(\sqrt{2},\sqrt[3]{4})$ and proceed from there, maybe first adjoining $\sqrt[3]{4}$, then obviously $\sqrt{2}$ is a root of $x^2 - 2$ so that could be the minimal polynomial of $\sqrt{2}$ over $\mathbb{Q}(\sqrt[3]{4})$ but first it should be proven that $\sqrt{2} \notin \mathbb{Q}(\sqrt[3]{4})$. This last step 'annoyed' me in this type of problems because I would just set $\sqrt{2}$ equal to an element of $\mathbb{Q}(\sqrt[3]{4})$ and go from there trying to show that the resulting equations have no solution which is cumbersome. So my question is, since we know that the degree of $\mathbb{Q}(\sqrt{2})$ over $\mathbb{Q}$ is 2 and the degree of $\mathbb{Q}(\sqrt[3]{4})$ over $\mathbb{Q}$ is 3, and these are contained in $\mathbb{Q}(\sqrt{2},\sqrt[3]{4})$ (illustrated below), can one not conclude immediately that the degree of this top extension over $\mathbb{Q}$ must be at least 6 by the tower theorem, and thus if either $\sqrt{2} \in \mathbb{Q}(\sqrt[3]{4})$ or $\sqrt[3]{4} \in \mathbb{Q}(\sqrt{2})$ we would get that $[\mathbb{Q}(\sqrt{2}, \sqrt[3]{4}):\mathbb{Q}] = 2$ or $[\mathbb{Q}(\sqrt{2}, \sqrt[3]{4}):\mathbb{Q}] = 3$ which is impossible? Or am I making a mistake somewhere?
For any field $K$ and any field extensions $K(\alpha),K(\beta)$ we have,
$$\operatorname{lcm}([K(\alpha):K],[K(\beta):K])\leq[K(\alpha,\beta):K]\leq[K(\alpha):K][K(\beta):K]$$
The upper bound follows fairly trivially from looking at bases for the associated vector spaces (to avoid making this answer too long I will avoid that here).
The lower bound follows from the tower law as,
$$[K(\alpha,\beta):K]=[K(\alpha,\beta):K(\alpha)][K(\alpha):K]$$ which implies $[K(\alpha):K]$ divides $[K(\alpha,\beta):K]$, and,
$$[K(\alpha,\beta):K]=[K(\alpha,\beta):K(\beta)][K(\beta):K]$$ which implies $[K(\beta):K]$ divides $[K(\alpha,\beta):K]$
Therefore $\operatorname{lcm}([K(\alpha):K],[K(\beta):K])$ divides $[K(\alpha,\beta):K]$ and so, $$\operatorname{lcm}([K(\alpha):K],[K(\beta):K])\leq[K(\alpha,\beta):K]$$
We can see from this that if $[K(\alpha):K]$ and $[K(\beta):K]$ are coprime then, $$\operatorname{lcm}([K(\alpha):K],[K(\beta):K])=[K(\alpha):K][K(\beta):K]$$ and so, $$[K(\alpha,\beta):K]=[K(\alpha):K][K(\beta):K]$$
$\blacksquare$
This applies directly to the case you are looking at and so we can see that $[\mathbb{Q}(\sqrt{2},\sqrt[3]{4}):\mathbb{Q}]=6$.