= Solution
Let $E=F(\alpha^p)\subseteq F(\alpha)$. If $\alpha$ is a <separable algebraic element> over $F$, its minimal polynomial over $E$ divides the original separable <minimal polynomial>. It also divides $T^p-\alpha^p=(T-\alpha)^p$, which has only one distinct root in any extension. Being separable with only one distinct root, that minimal polynomial has degree one. Hence $\alpha\in E$ and $F(\alpha)=F(\alpha^p)$.
Conversely, if the degree-$n$ minimal polynomial $m$ over $F$ is inseparable, part (b) writes it as $m(T)=g(T^p)$ with $\deg g=n/p$. Then $g(\alpha^p)=0$, so
$$
[F(\alpha^p):F]\leq n/p<n=[F(\alpha):F].
$$
The fields cannot be equal. This proves the equivalence
$$
\boxed{\alpha\text{ separable over }F\iff F(\alpha)=F(\alpha^p)},
$$
the <separability criterion using a pth power>.
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