I think it suppose to be really straight forward however I can't find a way to justify it. Let $p$ be an irreducible polynomial $x^3 + bx + c \in \mathbb{Q}[x]$ with $b>0$. Then by deriving it as a real function, one learn that it has only one real root, and two complex roots. so $p(x) = (x-r)(x^2+dx+e)$ when $r$ is $p's$ real root and $z,\bar{z}$ are the complex roots. I wish to show that $\bar{z}\in \mathbb{Q}(z)$ in order to claim that $[\mathbb{Q}(z,\bar{z}):\mathbb{Q}(z)]=2$ (and thus $[\mathbb{Q}(r,z,\bar{z}):\mathbb{Q}] = 3\times2)$.
What I tried to do was using $z = \frac{\bar{z}z}{\bar{z}}$ and the fact that $z^2+dz+e=0$, and thus $(\frac{\bar{z}z}{\bar{z}})^2+d\frac{\bar{z}z}{\bar{z}} +e=0\Rightarrow (z\bar{z})^2+d(\bar{z}^2)+e\bar{z}^2=0$ , but I don't find a way to go on.
Edit:
Well a counter example appears in comments. My intention was to demonstrate that the rank of the splitting field must be $6$ , so I'll explain a way around it (cause $\mathbb{Q}(z,\bar{z})\neq \mathbb{Q(z)} $ generally). In case $r$ is a real root, then $p(x)/(x-r)\in \mathbb{Q}(r)$ due to: $(x-r)(x^2+dx+e)=x^3+bx+c \Rightarrow x^3 + (d-r)x^2 + (e-rd)x -re = x^3+bx+c$ $\Rightarrow d=r , e=c/(-r)$ when $c\in \mathbb{Q}$.
Now $\mathbb{Q}(r)\subseteq \mathbb{R}$ and $x^2+dx+e$ is irreducible, as a polynomial of degree 2 without roots in $\mathbb{Q}(r)$ (complex roots). So, $[\mathbb{Q}(r)[x] / \langle x^2+dx+e\rangle :\mathbb{Q}(r)] = 2$ and $\mathbb{Q}(r)[x]/ \langle x^2+dx+e\rangle \cong (\mathbb{Q}(r))(z) = \mathbb{Q}(r,z)$ when $z$ is a root of $x^2+dx+e$.
A polynomial of degree 2 as a roots, thus it splits over $\mathbb{Q}(r,z)$ thus $\mathbb{Q}(r,z,\bar{z}) = \mathbb{Q}(r,z)$ and $\mathbb{Q}(r,z):\mathbb{Q} = [\mathbb{Q}(r,z):\mathbb{Q}(r)][\mathbb{Q}(r):\mathbb{Q}] = 2\times3 = 6$.
For the counter example with $x^3-2$ we get that $\mathbb{Q}(2^{1/3}, 2^{1/3}\omega) = \mathbb{Q}(2^{1/3}, 2^{1/3}\bar{\omega}) = \mathbb{Q}(2^{1/3}\omega, 2^{1/3} \bar{\omega})$