$\mathbb{Q}(\alpha, \sqrt 2)$ = $\mathbb{Q}(\alpha \sqrt 2)$, where $\alpha$ is a root of $ x^3 −x^2 + 1$? Also, $\mathbb{Q(\alpha, \sqrt 2)} = \mathbb{Q}(\alpha + \sqrt 2)$?
I know that the polynomial is irreducible over $\mathbb{Q}$, and also that to solve this exercise I must use the fact that $\alpha = -\alpha^4 + \alpha^2 + 1$.
Let $\beta=\alpha\sqrt{2}$.
It is clear that $\mathbb Q(\alpha\sqrt2)\subseteq \mathbb Q(\alpha,\sqrt2)$, so we will show the reverse inclusion.
We can rewrite the equation $\alpha=-\alpha^2+\alpha^2+1$ as:
$$\alpha=\frac{\beta^4}4+\frac{\beta^2}2+1$$
Thus $\alpha \in \mathbb Q(\beta)$. Then since $\sqrt{2}=\beta/\alpha \in \mathbb Q(\beta)$ as well, it follows that $\mathbb Q(\alpha,\sqrt2)\subseteq \mathbb Q(\alpha\sqrt2)$. Thus we are done.