Let $\mathbb{Q}(x)$ the smallest field extension of $\mathbb{C}/\mathbb{Q}$ containing $x\in \mathbb{C}$. Let $y\in \mathbb{Q}(x)$.
Also, $x$ is a root of $X^3-2X+2\in \mathbb{Q}[X]$ (doesn't need to be relevant)
Prove that $\mathbb{Q}(y) \in \{\mathbb{Q},\mathbb{Q}(x)\}$
I've already shown that the $\mathbb{Q}-$base of $\mathbb{Q}(x)$ is $\{1,x,x^2\}$.
Since $y = \lambda_0+\lambda_1 x + \lambda_2 x^2$, we know for $\lambda_1 \wedge \lambda_2 = 0 \Rightarrow \mathbb{Q}(\lambda_0) = \mathbb{Q}$.
So it remains to show that for $\lambda_1 \vee \lambda_2 \neq 0 \Rightarrow \mathbb{Q}(y) = \mathbb{Q}(x)$.
I'm not sure how to do that. I probably need to show that they have the same base? Anyways, I've made this attempt:
Since $$[\mathbb{Q}(y):\mathbb{Q}]=[\mathbb{Q}(x):\mathbb{Q}]\cdot [\mathbb{Q}(y):\mathbb{Q}(x)]=3\cdot [\mathbb{Q}(y):\mathbb{Q}(x)]$$ the dimension of $\mathbb{Q}(y)$ needs at least to be $3$. $\{1,x,x^2\}$ satisfies this criteria and $y$ can be displayed by this base by definition.
Is that enough?