Show that $f=x^3+7x+5$ has no roots in $\mathbb {Q}(\sqrt[4]{2})$

Question

Show that $f=x^3+7x+5$ has no roots in $\mathbb {Q}(\sqrt[4]{2})$.

I'm given a hint: suppose $\alpha$ is a root of $f=x^3+7x+5$ and $\alpha\in\mathbb{Q}(\sqrt[4]{2})$, compute $[{\mathbb{Q}(\alpha)}:\mathbb{Q}]$ and $[\mathbb{Q}(\sqrt[4]{2}):\mathbb{Q}]$.

This is what i got so far:

So if $\alpha$ is a root of f then $f(\alpha)=0$ which means $\alpha$ is algebraic over $\mathbb{Q}$, as f is nonzero monic and irreducible, $f=x^3+7x+5$ is irreducible in Q by RRT. This means that f is the minimal polynomial of $\alpha$ and as f has degree three $[\mathbb{Q}(\alpha):[\mathbb{Q}]]=3$. I also i know that the field extension of ${\mathbb{Q}(\sqrt[4]{2})}:[\mathbb{Q}]]$ is of degree 4. After this point im a little lost has to how to go about the problem, how do these facts (assuming what i have done is correct) imply there is no root in $\mathbb{Q}(\sqrt[4]{2})$??

Any pointers would be much appericated!

Answer 1

If $\mathbf{Q}(\alpha)$ is a subfield, then it's degree over $\mathbf{Q}$ must divide 4, as, there is the formula $[K:L]=[K:F][F:L]$ where $L \subseteq F \subseteq K$ are finite field extensions.(If $\alpha \in \mathbf{Q}(\sqrt[4]{2})$ then $\mathbf{Q}(\alpha)$ is a subfield of it)!

Answer 2

Hint : if $\alpha \in {\mathbb Q}[\sqrt[4]{2}]$ then $[{\mathbb Q}[\sqrt[4]{2}]:{\mathbb Q}]=[{\mathbb Q}[\sqrt[4]{2}]:{\mathbb Q}(\alpha)][{\mathbb Q}(\alpha):{\mathbb Q}]$