Showing $[\mathbb{Q}(\sqrt[4]{2},\sqrt{3}):\mathbb{Q}]=8$.

I think one can use reduction modulo $7$ to show that $\sqrt 3$ is not in $\mathbb{Q}[\sqrt[4]{2}]$, since in $\mathbb{F}_7$ the equation $X^4-2$ has a solution (i.e.\ $x=5$) but $X^2-3$ has no solution.

Edit: Easier proof: Assume that $\sqrt{3}\in \mathbb{Q}(\sqrt[4]{2})$. Then $\mathbb{Q}(\sqrt{2},\sqrt{3})\subseteq \mathbb{Q}(\sqrt[4]{2})$ and from the equality of degrees over $\mathbb{Q}$ those are equal. But this implies that $\mathbb{Q}(\sqrt[4]{2})$ is Galois which is contradiction, since $e^{2\pi i/4} \sqrt[4]{2}\not\in \mathbb{Q}(\sqrt[4]{2})$.

Here is an alternative method to show that $\mathbb{Q}(\sqrt{2},\sqrt{3}) = \mathbb{Q}(\sqrt[4]{2})$ results in a contradiction, without using any knowledge of Galois extensions (since the problem is asked in the textbook before this topic is covered).

A basis for $\mathbb{Q}(\sqrt{2},\sqrt{3})$ over $\mathbb{Q}$ is $\{ 1,\sqrt{2},\sqrt{3},\sqrt{6} \}$. By assumption, $\sqrt[4]{2} \in \mathbb{Q}(\sqrt{2},\sqrt{3})$, so there exist $a,b,c,d \in \mathbb{Q}$ such that $$ \begin{align} & & \sqrt[4]{2} &= a + b\sqrt{2} + c\sqrt{3} + d\sqrt{6}\\ &\implies &\sqrt{2} &= (a^2 + 2b^2 + 3c^2 + 6d^2) + 2(ab + 3cd)\sqrt{2} + 2(ac + 2bd)\sqrt{3} + 2(ad+bc)\sqrt{6}\\ &\implies &0 &= \color{blue}{(a^2 + 2b^2 + 3c^2 + 6d^2)} + 2(ab + 3cd - 1)\sqrt{2} + 2(ac + 2bd)\sqrt{3} + 2(ad+bc)\sqrt{6} \end{align} $$ Since $\{1,\sqrt{2},\sqrt{3},\sqrt{6}\}$ is a $\mathbb{Q}$-linearly independent set, each of the coefficients must be zero. In particular, $$ a^2 + 2b^2 + 3c^2 + 6d^2 = 0 \implies a=b=c=d=0 \implies \sqrt[4]{2} = 0, $$ which is a contradiction.