Let $F$ be a field and $a\in F$ and let $m,n$ be coprime positive integers such that. Then $x^{mn}−a$ is irreducible in $F[x]$ if and only if both $x^m−a $ and $x^n−a$ are irreducible in $F[x]$. This is supposed to be a Galois Theory question however I don't see how I can use Galois theory to prove such a result. I mean all I can say is that is that the splitting field of $x^{mn}-a$ is $F(z_n,c)$ where $z$ is a primitive nth root of unity and $c^n=a$ I have no idea how to continue from there any help/solution will be greatly appreciated.
Thanks in advance
This is an exercise on irreducibility, not on splitting fields. If $X^{mn}-a$ is irreducible over $F$, so are $X^{m}-a$ and $X^{n}-a$ because $X^{mn}-a=(X^{m})^n-a=(X^{n})^m-a$.
Conversely, assume $X^{m}-a$ and $X^{n}-a$ irreducible over $F$, and let $\alpha$ be a root of $X^{mn}-a$ in an algebraic closure. Then $\alpha^m$ is a root of $X^{n}-a$, hence $[F(\alpha ^m) :F]= n$, and similarly $[F(\alpha ^n) :F]= m$. If $m$ and $n$ are coprime, a classical result on the multiplicativity of degrees in towers ensures that $[F(\alpha ^m, \alpha ^n):F]= mn$. But by Bezout's theorem, there exist integers $r, s$ s.t. $rm+sn=1$, so that $\alpha = \alpha^{rm} \alpha^{sn}$, and $F(\alpha ^m, \alpha ^n)=F(\alpha)$. Hence the degree of $\alpha$ over $F$ is $mn$, and $X^{mn}-a$ must be irreducible over $F$.