# Irreducible cubic polynomial in $\mathbb{Q}[x]$ has no roots in $\mathbb{Q}(\sqrt{2}, 5^{1/4})$

I am working on the following problem:

Consider the field extension $$F = \mathbb{Q}(\sqrt{2}, 5^{1/4})$$ of $$\mathbb{Q}$$. Let $$f(x)$$ be an irreducible cubic polynomial in $$\mathbb{Q}[x]$$. Prove that $$F$$ contains no roots of $$f(x)$$.

I know that, since $$f(x)$$ an irreducible degree $$3$$ polynomial in $$\mathbb{Q}[x]$$, $$f(x)$$ has no roots in $$\mathbb{Q}$$. In the specified extension $$F$$ of $$\mathbb{Q}$$, we're only adjoining $$\sqrt{2}$$ and $$5^{1/4}$$. Thus, it's left to show that $$f(x)$$ can't have a root that is a power of $$\sqrt{2}$$ or a power of $$5^{1/4}$$.

I think I can see how to partially argue this. I believe $$f(x)$$ can't have a root that is an odd power of $$5^{1/4}$$, right off the bat, because the corresponding minimal polynomial for such an element would have to be degree $$4$$, already contradicting the degree of $$f(x)$$. Thus, the problem has now reduced down to showing that $$f(x)$$ can't have a root that is a power of $$\sqrt{2}$$ or a power of $$\sqrt{5}$$. But I'm not sure how to argue these remaining two points. Does it have to do with the minimal polynomial of such an element being of degree $$2$$, which would need to be a factor of $$f(x)$$ ? And this contradicts $$f(x)$$ being irreducible?

Any help would be appreciated. Thanks!

If $$F$$ does contain a root $$r$$ of $$f$$, then we would have a field embedding $$K = \mathbb Q[x]/f(x) \hookrightarrow F$$, sending $$x$$ to $$r$$.
We then have a problem with the dimensions: $$\dim_\mathbb Q F = \dim_\mathbb Q K \cdot \dim_K F$$. It should be clear that $$\dim_\mathbb Q F = 8$$, but $$\dim_\mathbb Q K = 3$$, so that's not possible.