Question :
Let $\alpha$ be a root of $x^3+x+1$ and $\beta$ be a root of $x^3+x+3$. Show that it is not possible that $\alpha\in \mathbb Q(\beta)$.
My proof :
Given $\beta$ is a root of $x^3+x+3$.
Thus $\beta^3+\beta+3 = 0$. Then $(\beta^3+\beta+2 ) + 1 = 0$.
If $\alpha\in\mathbb Q(\beta)$, then $\alpha = r_1 + r_2\beta$ for some rationals $r_1$ and $r_2$.
Also $(r_1+r_2\beta)^3+r_1+r_2\beta+1 = 0$.
Therefore for some rationals $r_1$ and $r_2$ we have $(r_1+r_2\beta)^3+r_1 + r_2\beta = \beta^3+\beta+2$.
But equating and solving such $r_1$ and $r_2$ doesn't exist.
Thus, $\alpha$ doesn't belong to $\mathbb Q(\beta)$.
Do you think my proof is right?
If not correct me or provide a better easier proof
Let $\alpha$ be a root of $x_3+x+1$ and $\beta$ be a root of $x_3+x+3$, then $$[\mathbb{Q}(\alpha):\mathbb{Q}]=[\mathbb{Q}(\beta):\mathbb{Q}]=3.$$ If $\alpha\in \mathbb{Q}(\beta)$, we know $\mathbb{Q}(\alpha)\subseteq \mathbb{Q}(\beta)$, so $\mathbb{Q}(\alpha)=\mathbb{Q}(\beta)$.
the fact: Let $L$ be a number field, and we assume that $\mathcal{O}_L$ is the ring of algebraic integers in $L$. Suppose that $\alpha_1,\cdots,\alpha_n$ is the elements of $\mathcal{O}_L$, then $\alpha_1,\cdots,\alpha_n$ is a integral basis of $L$ iff $d_L(\alpha_1,\cdots,\alpha_n)$ has no square factors.
We have $d_{\mathbb{Q}(\alpha)}(1,\alpha,\alpha^2)=-31$, so $\{1,\alpha,\alpha^2\}$ is a integral basis of $\mathbb{Q}(\alpha)$ by using the fact. Thus we can get the discriminant of $\mathbb{Q}(\alpha)$: $$d(\mathbb{Q}(\alpha))=d_{\mathbb{Q}(\alpha)}(1,\alpha,\alpha^2)=-31.$$
In the similar way, we know that $d_{\mathbb{Q}(\beta)}(1,\beta,\beta^2)=-247=-13\times 19$ has no square factors, thus $$d(\mathbb{Q}(\beta))=d_{\mathbb{Q}(\beta)}(1,\beta,\beta^2)=-247\ne d(\mathbb{Q}(\alpha)).$$
That is, $\mathbb{Q}(\alpha)\ne \mathbb{Q}(\beta)$. So $\alpha \notin \mathbb{Q}(\beta)$.