How to show that $f(x)=x^3+\omega x+1$ is irreducible over $\Bbb C(\omega)$? This is what I have tried:
Suppose on the contrary that: $$f(x)=g(x)h(x)=(x-\alpha(\omega))(x^2+\beta(\omega)x+\gamma(\omega)).$$ Then $\alpha(\omega)\gamma(\omega)=-1$ and $\gamma(\omega)-\alpha(\omega)\beta(\omega)=\omega$ and $\alpha(\omega)=\beta(\omega)$.
Now, I should conclude that this case is not possible, but I can't see why. Where is contradiction? Thanks in advance.
Suppose there exist elements $\,a\,,\,b\,,\,c\in\Bbb C(w)\;$ s.t.
$$x^3+wx+1=(x-a)(x^2+bx+c)=x^3+(b-a)x^2+(c-ab)x-ac\implies$$
$$\begin{align*}a&=b\\ ac&=-1\\ c-ab&=c-a^2=w\end{align*}$$
But since the second equality above means $\,a,c\,$ are non-zero polynomials in $\,w\,$ of degree zero (i.e., constants or simply non-zero complex numbers), the last equality tells us that in fact $\,w\in\Bbb C\,$ , which is absurd.