Degree of $\mathbb{Q}(\sqrt[4]{3},\sqrt[5]{3})$ over $\mathbb{Q}$

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Call $d$ the degree I want to find.

I know that

$d=[\mathbb{Q}(\sqrt[4]{3},\sqrt[5]{3}):\mathbb{Q}(\sqrt[4]{3})][\mathbb{Q}(\sqrt[4]{3}):\mathbb{Q}]=[\mathbb{Q}(\sqrt[4]{3},\sqrt[5]{3}):\mathbb{Q}(\sqrt[5]{3})][\mathbb{Q}(\sqrt[5]{3}):\mathbb{Q}]$

$[\mathbb{Q}(\sqrt[4]{3}):\mathbb{Q}]=4$

$[\mathbb{Q}(\sqrt[5]{3}):\mathbb{Q}]=5$

So $4$ and $5$ divides $d$

My guess is that $d=20$ so I'm trying to prove that $f=x^5-3$ is irreducible over $\mathbb{Q}(\sqrt[4]{3})$, but I'm stuck at it.

Any tips?