$\beta \in \mathbb{Q}(\alpha)$ 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!
 A: 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 $\Bbb Q(\alpha) = \Bbb Q(\beta)$.
A: How to tell that two irreducible polynomials do not produce isomorphic extensions:
Let $f$, $g$ be irreducible polynomials in $k[x]$ ( of same degree) that produce isomorphic field extensions $k[x]/(f)$, and   $k[x]/(g)$. Then for their discriminants $\Delta(f)$, $\Delta(g)$, we have
$$\frac{\Delta(f)}{\Delta(g)} \in (k^{\times})^2$$
Now, the discriminant of the polynomial of the polynomial $x^3- b$ is $-27 b^2$, while the discriminant of a polynomial of degree $3$ with all real roots is a positive number.
$\bf{Added:}$ Let's show the statement about discriminants of polynomials. Consider a polynomial $f\in k[x]$ with roots $x_1$, $\ldots$, $x_n$, and $g\in k[x]$ with roots $t(x_1)$, $\ldots$, $t(x_n)$, where $t(x) \in k(x)$ is a rational function. Then the expression in $x_i$
$$\prod_{i<j} \frac{t(x_i)- t(x_j)}{x_i-x_j}$$
is symmetric, so lies in $k$. Therefore, the quotient of the discriminants of $g$ and $f$ is the square of an element from $k$.
A: Here is another approach. We have $$\beta=p+q\alpha +r\alpha^2$$ for some rationals $p, q, r$. Replacing $\alpha $ in above equation with the other two roots of $f$ we get two real numbers $\beta _1,\beta_2$ and then by the theorem on symmetric polynomials $\beta, \beta_1,\beta_2$ are roots of a cubic polynomial $g(x) \in\mathbb Q[x] $. And since $\beta$ is of degree $3$ over rationals the polynomial $g$ is irreducible over $\mathbb {Q} $. Also $\beta^3$ is rational so the above polynomial $g(x) $ must be of the form $ax^3+b$.
But then we reach a contradiction as the polynomial $ax^3+b$ has at least one root which is not real.
