$X^m-a$ and $X^m-b$ irreducible with same splitting field, prove relation between $a$ and $b$.

Let $K$ be a field containing a primitive $m$-th root of unity $\zeta_m$ and suppose $a,b\in K$ are such that $X^m-a$ and $X^m-b$ are irreducible polynomials over $K$ that have the same splitting field. I'm meant to show there exist $c\in K$ and $r\in\mathbb{N}$ such that $r,m$ are coprime and $b=c^ma^r$.

My thoughts: Note that as $\zeta_m\in K$ by assumption, a splitting field for $X^m-a$ is $K(\alpha)$ where $\alpha^m=a.$ Knowing $X^m-b$ splits over $K(\alpha)$ then means $\exists\beta\in K(\alpha)$ such that $\beta^m=b,$ say $\beta=c_0+c_1\alpha+\dots+c_{m-1}\alpha^{m-1}.$ Looking at the relation I'm aiming for, it would be nice if I could deduce somehow that all but one of the $c_i$ are zero, using $\beta^m=b\in K.$ I'm not quite seeing how to do that though.

Of course, a different approach would also be welcome.

Side note - showing the converse is rather straightforward. That is, given $b=c^ma^r$, the splitting fields are the same.

Just so I have something to accept; as suggested by reuns, $Gal(K(\alpha)/K)\cong C_m$ and is generated by a $K$-automorphism $\sigma$ mapping $\alpha\mapsto\zeta_m\alpha.$ Being a $K$-automorphism, $\sigma$ must map $\beta$ to a root of $X^m-b$, say $\zeta_m^r\beta.$

However, not any old $r$ will do - note that $\sigma^s$ is not the identity map for any $s<m.$ This immediately means $\sigma^s$ doesn't fix $\beta$ for any $s<m$, as a $K$-automorphism which fixes $\beta$ must fix all of $K(\beta)=K(\alpha).$ Thus $r$ must be coprime to $m$. So take $j$ to be the inverse of $r$ mod $m$ and then $\sigma(\beta^j)=\zeta_m\beta^j.$

So then $\sigma$, which of course only fixes $K$ (as any power of $\sigma$ also fixes what $\sigma$ does, so $\langle\sigma\rangle=Gal(K(\alpha)/K)$ fixes it) sends $\alpha\mapsto\zeta_m\alpha$ and $\beta^j\mapsto\zeta_m\beta^j$, i.e $\sigma$ fixes $\alpha/\beta^j.$ So $\alpha/\beta^j=c\in K$ and raising to the $m$th power gives the result.