$\beta \in \mathbb{Q}(\alpha)$ such that $\mathbb{Q}(\alpha) = \mathbb{Q}(\beta)$ and $\beta^3 \in \mathbb{Q}$ does not exist. I'm stuck on proving that.
Let $f \in \mathbb{Q}[X]$ is irreducible cubic polynomial which has three real roots, and $f(\alpha)=0$,
then there does not exist $\beta \in \mathbb{Q}(\alpha)$ such that $\mathbb{Q}(\alpha) = \mathbb{Q}(\beta)$ and $\beta^3 \in \mathbb{Q}$.
I'll write down what I noticed.

*

*$1, \alpha, \alpha^2$ is base of $\mathbb{Q}(\alpha)$

*Let $g$ is minimal polynomial of $\beta$ on $\mathbb{Q}$, $\mathrm{deg}(g) = 3$.

*Let $x, y, z \in \mathbb{R}$ are roots of $f$ ($\alpha$ is one of them), then $xyz, x+y+z, xy+xz+yz \in \mathbb{Q}$
please some hint or solution.
Thank you!

Suppose there exists such a $\beta$. It is clear that $\beta \not \in \Bbb Q$.
Then there is an embedding $\Bbb Q(\beta) \hookrightarrow \Bbb C$ which maps $\beta$ to a non-real complex number, because the three roots of $X^3 - \beta^3$ are not all real.
However since $f(X)$ has three real roots, any embedding $\Bbb Q(\alpha) \hookrightarrow\Bbb C$ would have image contained in $\Bbb R$. This contradicts $\Bbb Q(\alpha) = \Bbb Q(\beta)$.
