I know the polynomial is irreducible over $\Bbb Q$ and the roots are real irrational. A $\Bbb Q$-automorphism $\sigma:\Bbb Q(\alpha)\to \Bbb Q(\alpha)$ is determined by $\sigma(\alpha)$, which is also a root of the polynomial. My problem is finding out if the other roots are also in $\Bbb Q(\alpha)$. If the other roots are $\beta, \gamma$, then I set the system $3=\alpha+\beta+\gamma, -3=\alpha\beta\gamma$. For example, clearing $\beta$, I obtain
$\beta=\frac{3-\alpha}{2}\pm\sqrt{\alpha^2-6\alpha+9+\frac{12}{\alpha}}$
(A solution can be with the positive square root, then the negative one is $\gamma$)
I don't know how to conclude if this $\beta$ is in $\Bbb Q(\alpha)$ or not.
I tried to search similar problems but they apply results I don't know, like using the discriminant or splitting stuff. With the current theory I have in lessons I only have very basics tools.
This is pure cheating: I browsed the referenced links and transformed their answers to fit this problem:
$$ P(x) = x^3 - 3x^2+3 $$
$$ P(\alpha) = 0 $$
$$ \beta = R(\alpha) = -\alpha^2 + \alpha + 3 $$
$$ P(\beta) = P (R(\alpha) = P ( -\alpha^2 + \alpha + 3) $$ $$ = - \alpha^{6} + 3 \alpha^{5} + 3 \alpha^{4} - 11 \alpha^{3} - 3 \alpha^{2} + 9 \alpha + 3 = $$ $$ = \left( \alpha^{3} - 3 \alpha^{2} + 3 \right) \left( -\alpha^{3} + 3 \alpha + 1 \right) = P(\alpha) \left( -\alpha^{3} + 3 \alpha + 1 \right) = 0 $$
Which means that $\beta$ is a root of $P$ and $\beta \in \mathbb Q(\alpha)$