Show that $\sqrt[15]{63}$ does not belong in $\mathbb Q(\sqrt[189]{147})$

Question

How to prove that $\sqrt[15]{63}$ does not belong in $\mathbb Q(\sqrt[189]{147})$?

Answer 1

If $\beta \in \mathbb Q (\alpha)$, then $\mathbb Q (\beta) \subseteq\mathbb Q (\alpha)$ and so $[\mathbb Q (\alpha): \mathbb Q] = [\mathbb Q (\alpha): \mathbb Q (\beta)]\cdot[\mathbb Q (\beta):\mathbb Q]$.

$\mathbb Q (147^{1/189})$ has degree $189$ over $\mathbb Q$.

$63^{1/15}$ has degree $15$ over $\mathbb Q$.

$15$ does not divide $189$.