Let $f\in \Bbb Q [x]$ be monic irreducible and of degree 3. Prove that if $\alpha \in \Bbb C$ is any zero of $f$ then the field $\Bbb Q(\alpha, \sqrt{\vartriangle(f)}) \subset \Bbb C$ , is a splitting field of $f$. It's neccesary that $f$ is irreducible?
Well I can start to do this problem in a very computationally way (without success) . Let's call the roots of $f$ by $ a,b,c$. We want to show that $b,c \in K=\Bbb Q(a,\sqrt{\vartriangle(f)}) $
Where $ \sqrt{\vartriangle(f)}=(a-b)(a-c)(b-c)=u$
Well since $(x-a)(x-b)(x-c) = x^3 - x^2(a+b+c)+x(ab+ac+bc)-abc \in \Bbb Q[x]$ We can conclude in particular that $abc \in \Bbb Q \subset K$ but since $a \in K$ then $ bc \in K$ . And also $a+b+c , a \in K$ implies that $b+c \in K$. Until now I never used that I have $u\in K$. Well I'll start to expand $ u= a^2b-a^2c + b^2c-b^2a+c^2a-c^2b \in K$. Well now maybe with some tricks I can deduce that $b,c \in K$ But i can't see it.
Please someone helpme with this computationally proof of this. And if someone has an argument using ideas of splitting fields and Galois theory are also welcome