Quick way of showing $[\mathbb{Q}(\sqrt[4]2,\sqrt3):\mathbb{Q}]=8$? 
Show that $[\mathbb{Q}(\sqrt[4]2,\sqrt3):\mathbb{Q}]=8$.

Clearly $[\mathbb{Q}(\sqrt[4]2):\mathbb{Q}]=4$, so it suffices to show $\sqrt3\notin\mathbb{Q}(\sqrt[4]2)$. I could only think of setting $(a+b\sqrt[4]2+c\sqrt[4]4+d\sqrt[4]8)^2=(e\sqrt3)^2$ and calculate, but this is tedious. Is there any quick way to do that?
 A: *

*Prove the only quadratic subfield of $\mathbf Q(\sqrt[4]{2})$ is $\mathbf Q(\sqrt{2})$ (think about field automorphisms), so $\sqrt{3} \not\in \mathbf Q(\sqrt[4]{2})$ since $\sqrt{3} \not\in \mathbf Q(\sqrt{2})$.

*If you know some algebraic number theory,  a quick solution is that $2$ is the only ramified prime number in $\mathbf Q(\sqrt[4]{2})$, so $3$ ramifying in $\mathbf Q(\sqrt{3})$ implies this quadratic field is not contained in $\mathbf Q(\sqrt[4]{2})$.

*The ring $\mathbf Z[\sqrt{3}]$ is Euclidean and thus is a UFD. In this  ring $2$ is not prime:  $2 = (\sqrt{3}+1)(\sqrt{3}-1)$, with both factors being prime since the absolute values of their norms in $\mathbf Z$ are prime. Therefore $x^4 - 2 = x^4 - (\sqrt{3}+1)(\sqrt{3}-1)$ is Eisenstein at a prime in $\mathbf Z[\sqrt{3}]$, so it is irreducible over the fraction field $\mathbf Q(\sqrt{3})$, which means adjoining a root of it to $\mathbf Q(\sqrt{3})$ creates a $4$th-degree extension of that quadratic field. Thus 
$[\mathbf Q(\sqrt{3},\sqrt[4]{2}):\mathbf Q] = 8$.

*In the column of "Related" questions to this one, I see your question was already asked a few years ago here, and on that page Lior Bary-Soroker provides yet another solution: if $\sqrt{3} \in \mathbf Q(\sqrt[4]{2})$ then $\mathbf Q(\sqrt{2},\sqrt{3}) = \mathbf Q(\sqrt[4]{2})$ since the left side is in the right side and the two fields have equal degree over $\mathbf Q$, but this equality is not possible since the field on the left is Galois over $\mathbf Q$ and the field on the right is not.
