 Solution

Let $\beta$ satisfy ${\beta }^p=\alpha$. Since $K=\mathbb{Q}(\alpha )\subseteq \mathbb{Q}(\beta )$,The polynomial $f(X^p)$ has degree $p[K:\mathbb{Q}]$ and has root $\beta$. It is irreducible exactly when $[K(\beta ):K]=p$.

If $\alpha ={\gamma }^p$ for some $\gamma \in K$, then $\gamma$ is a root of $f(X^p)$ of degree at most $[K:\mathbb{Q}]$, so $f(X^p)$ is reducible. Conversely, if it is reducible, thenSince $p$ is prime, $1\le d<p$ implies $\gcd (d,p)=1$. Applying coprime-degree descent for powers to $K(\beta )/K$ and ${\beta }^p=\alpha$ shows that $\alpha$ is a $p$th power in $K$. Hence