$\Bbb Q(a, \sqrt{\vartriangle(f)})$ is a splitting field in this case.

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

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Since $f$ is irreducible of degree $3$ we get that $[\mathbb{Q}(\alpha):\mathbb{Q}]=3$ and that its Galois group is either cyclic of order $3$, i.e. $C_3 (=A_3)$, or the full symmetric group $S_3$ (because these are all the transitive subgroups of $S_3$).
In the first case, $\mathbb{Q}(\alpha)$ is already the splitting field of $f$. Since the discriminant is a square iff the Galois group consists of even permutations, we also get that $\sqrt{\Delta}\in \mathbb{Q}$, so $\mathbb{Q}(\alpha)=\mathbb{Q}(\alpha,\sqrt{\Delta})$.
In the second case, $[\mathbb{Q}(\alpha,\sqrt{\Delta}):\mathbb{Q}]=6=|S_3|$, since $2=[\mathbb{Q}(\sqrt{\Delta}):\mathbb{Q}]$ and $3=[\mathbb{Q}(\alpha):\mathbb{Q}]$ are coprime. Thus $\mathbb{Q}(\alpha,\sqrt{\Delta})$ is a subfield of the splitting field, and of the same degree, hence equals the splitting field.