= Solution
Put $\gamma=\alpha+\beta$, $E=F(\gamma)$ and $E_0=F(\gamma^p)$. By the preceding <separability criterion using a pth power>, it suffices to prove $E=E_0$.
Since $\alpha$ is separable, $\alpha\in F(\alpha^p)$. But $\alpha^p=\gamma^p-\beta^p\in E_0(\beta)$, so $\alpha\in E_0(\beta)$. Consequently
$$
M:=F(\alpha,\beta)=E(\beta)=E_0(\beta).
$$
Let $h$ and $m$ be the monic <minimal polynomials> of $\beta$ over $E_0$ and $E$, respectively. Both divide its separable minimal polynomial over $F$, so both are separable. Also $m\mid h$ in $E[T]$.
Every $c\in E=F(\gamma)$ has $c^p\in F(\gamma^p)=E_0$. Hence $m(T)^p$ has coefficients in $E_0$ and vanishes at $\beta$, forcing $h\mid m^p$ in $E_0[T]$. Since $h$ remains a <square-free polynomial> over $E$, this divisibility implies $h\mid m$ in $E[T]$. Together with $m\mid h$, monicity gives $h=m$. Therefore
$$
[M:E_0]=\deg h=\deg m=[M:E].
$$
The <tower law> makes $[E:E_0]=1$. Part (c) now proves that $\gamma=\alpha+\beta$ is separable over $F$. This establishes <closure of separable algebraic elements under addition> without presupposing that closure as a field theorem.
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