There exist $y \in K$ and $b \in E \setminus F$ such that $\min(y,E)(X) = X^3-b$ in a concrete fields $F \leq E \leq K$

Question

Let $F \leq E \leq K$ be a finite extension of fields, with $[K : F] = 6$ and $[E : F] = 2$ (so $[K : E] = 3$), such that $K / E$ is normal and $E = F(x)$, being $x \in E$ a root of $X^2 + X + 1$. I have to show that there exist $y \in K$ and $b \in E \setminus F$ such that $\min(y , E)(X) = X^3 - b$ (I can use also that the characteristic of $F$ is neither $2$ nor $3$ if I needed). My attempt is the next: the extension $K / F$ is finite and I have shown that it is separable too, so using the primitive element theorem, there exists $y \in K$ such that $K = E(y)$. Generally, using that $E = F(x)$, we have that there exist ${\alpha}_i , {\beta}_i \in F$ ($i \in \{0 , 1 , 2\}$) such that $$ \min(y , E)(X) = X^3 + ({\alpha}_2 + {\beta}_2 x) X^2 + ({\alpha}_1 + {\beta}_1 x) X + {\alpha}_0 + {\beta}_0 x $$ as $3 = [K : E] = \deg \min(y , E)$ and the elements of $E$ are like $\alpha + \beta x$, with $\alpha , \beta \in F$. We have that $K / E$ is a Galois extension too, as it is normal, but it is also separable (as $K / F$ is separable). How can I use it? In fact, $E / F$ is a Galois extension too, because it is normal (as each any cuadratic extension is normal) and it is separable (as $K / F$ is separable). I know how I can use neither this nor that $x$ is a primitive cubic root of unity to conclude that $\min(y , E)(X) = X^3 - b$, with $b \in E \setminus F$. Thank you very much in advance.

Answer 1

This is one of the elementary lemmas leading to Kummer theory.

You know that $K/E$ is Galois with Galois group cyclic of order three. If $\sigma$ is a generator of the Galois group, and $\omega\in E$ is a zero of $x^2+x+1$, then Artin's Lemma on independence of characters gives that there exists an element $z\in E$ such that $$x:=z+\omega\sigma(z)+\omega^2\sigma(z)\neq0.$$ Attack:

1. Show that $\sigma(x)=\omega^{-1}x$. Conclude that $x\notin E$.
2. Show that $\sigma(x^3)=x^3$. Conclude that $x^3\in E$.