We recently completed seprability from Cohn's book (Basic Algebra) and I am stuck on this question.
I was thinking along the lines that let $f\in K[x]$ be the irreducible polynomial where $K$ is a field , $\alpha$ and $\beta$ be two of its zeroes in some extension with multiplicity $m$ and $n$ resp. Then $f(x)=(x-\alpha)^mg(x)$ in $K(\alpha)[x]$ and $f(x)=(x-\beta)^nh(x)$ in $K(\beta)[x]$ where $g(\alpha)\neq0$ and $h(\beta)\neq0$.
Since $f$ is irreducible we have a $K$- isomorphism $\sigma: K(\alpha) \to K(\beta)$ which takes $\alpha \to \beta$. From here we should be able to do something to get $m=n$ or assume them to be unequal and then get a contradiction. But I could not think of any way. Any help here is appreciated. Thanks.
$f^{\sigma}=(x-\beta)^mg^{\sigma}(x)$ implies that $m\leq n$, by using $\sigma^{-1}$, we deduce that $n\leq m$.