Let $F \subseteq E \subseteq K$ a chain of fields, such that the characteristic of $F$ is neither $2$ nor $3$, and we suppose that we have found $z \in K$, with $K = E(z)$, and $a \in E$ such that $\min(z , E)(X) = X^3 - a$.
We suppose also that $E = F(x)$, with $[E : F] = 2$, such that $x$ is a root of the polynomial $X^2 + X + 1$ (so $x$ is a primitive cube root of unity). $K / E$ and $E / F$ have to be Galois extensions. Can we take $a$ such that $a \in F$? I only look for an example.
I thought in $F = \mathbb{Q}$, $E = F(e^{2 \pi i/3})$ and $K = E(\sqrt[3]{2})$ and $a = 2$, I am not sure if $K / E$ is normal though.
In general you cannot take $a\in F$; your example is close to a counterexample:
Take $F = \mathbb{Q}$ and $E = F\left(e^{\frac{2 \pi i}{3}}\right)$, and $$K=E\left(e^{\frac{2 \pi i}{9}}\right).$$ Then we may take $z=e^{\frac{2 \pi i}{9}}$, but $\min(z,E)=X^3-e^{\frac{2 \pi i}{3}}$.
In fact, for any choice of non-cube $a\in\mathbb{Q}$, adjoining a root $z$ of $X^3-a$ to $E=\mathbb{Q}(e^{\frac{2 \pi i}{3}})$ yields a splitting field of $X^3-a$, which has Galois group $S_3$, whereas $K=\mathbb{Q}(e^{\frac{2 \pi i}{9}})$ has Galois group $\mathbb{Z}/6\mathbb{Z}$.