Prove that degree of a special field extension is $3$ or $6$.

Question

There is a field $K$, where $\text{char} K = 0$

We have an irreducible cubic polynomial $f(x)=ax^3+bx^2+cx+d\in K[X]$.

The polnomial has following roots: $x_1, x_2, x_3$

Let $K(x_1,x_2,x_3)$ be an extension field

Prove that $[K(x_1,x_2,x_3): K]$ is equal to $3$ or $6$

I know that $[K(x_1,x_2,x_3): K] | (\deg f)!$ Where $(\deg f)!=6$,

but why $[K(x_1,x_2,x_3): K]$ is different from $1,2$ ?

Let $\Delta:= a^4(x_1-x_2)^2(x_2-x_3)^2(x_3-x_1)^2$ be a discriminant.

Additionally i would like to know why if $\sqrt{\Delta}\in K$ then $[K(x_1,x_2,x_3): K]=3$ and if $\sqrt{\Delta}\notin K$ then $[K(x_1,x_2,x_3): K]=6$

Regards

Answer 1

Hint: Show that $[K(x_1):K]=3$, $[K(x_1,x_2):K(x_1)]\leq 2$, $[K(x_1,x_2,x_3):K(x_1,x_2)]\leq 1$.