$F$ is a field and we have $0\neq n\geq 2$. I need to show that $f=x^n-a$ is separable, meaning all its irreducible factors are separable.
I first tried induction, but without avail. I couldn't figure out how to factor out $x^n-a$ and use the induction hypothesis.
Then I tried to look at $f$ when it is irreducible and reducible. It is trivial when $f$ is irreducible, and when it's reducible I let $f=p_1p_2\dots p_n$ but I don't know how to proceed from here...
I learned about the equivalence between $f$ having derivative not equal to zero and $f$ not having multiple roots in any field extension of $F$ and I wanted to use this theorem but I can't figure out how.
As was mentioned in the comments, there are a number of ways to possibly show $f(x) := x^n-a$ is separable. The most common are either to show that $f$ and $f’$ are coprime or to show that $f$ has $n$ distinct roots. I’ll do the later.
We will show that $\zeta_n^i \sqrt[n]{a}$ are the distinct roots of $f$, for $i \in \{0,...,n-1\}$, where $\zeta_n$ is a primitive $n^{th}$ root of unity.
Recall that, for $i,j \in \{0,...,n-1\}$, $i \not = j$, $\zeta_n^i \not = \zeta_n^j$, so the $\zeta_n^i \sqrt[n]{a}$ are distinct. To show these are roots of $f$, consider $f(\zeta_n^i \sqrt[n]{a})$.
$$f(\zeta_n^i \sqrt[n]{a}) = (\zeta_n^i \sqrt[n]{a})^n - a = (\zeta_n^i)^n (\sqrt[n]{a})^n - a = (\zeta_n^n)^i (a) - a = (1)^i(a) - a = a-a = 0$$
Therefore, for $i \in \{0,...,n-1\}$, $\zeta_n^i \sqrt[n]{a}$ is a root of $f$. Thus, $f$ has $n$ distinct roots, so $f$ is separable.