I assume it is given that there are real roots, else the answer is trivially no. If I don't have calculation mistakes, this should do it. Denote $ r_{1}=a,r_{2}=b,r_{3}=c$ for simplicity, then we have $ a-b,a+b+c,ab+bc+ca,abc\in\mathbb{Q}$ by de Viete. So, $ c^{2}+2ab=a^{2}+b^{2}+c^{2}-(a-b)^{2}=(a+b+c)^{2}-2(ab+bc+ca)-(a-b)^{2}\in\mathbb{Q}$, and $ c^{2}-2ac-2bc\in\mathbb{Q}\Rightarrow c(c-2a-2b)\in\mathbb{Q}$. Now, $ c(a_{1}-3(a+b))\in\mathbb{Q}\Rightarrow c(a_{1}+3c)=3c^{2}+ca_{1}\in\mathbb{Q}\Rightarrow c^{2}+\frac{ca_{1}}{3}\in\mathbb{Q}$. Since we had $ c^{2}+2ab\in\mathbb{Q}$ this gives $ 6ab-ca_{1}\in\mathbb{Q}\Rightarrow6ab-c(a+b+c)\in\mathbb{Q}$, adding the rationals $ c^{2}+2ab$ and $ ab+bc+ca$ this yields $ 9ab\in\mathbb{Q}$, so $ ab\in\mathbb{Q}$. And then $ c^{2}\in\mathbb{Q}$, so $ a_{1}^{2}+ab-a_{1}(a+b)=(a_{1}-b)(a_{1}-a)=(c+a)(c+b)=c^{2}+ab+bc+ca\in\mathbb{Q}$, meaning $ a+b\in\mathbb{Q}$. Together with $ a-b\in\mathbb{Q}$, this solves the problem.

You mean $ a_{1}=a+b+c$?

Sorry, I thought $ p(x)=x^{3}-a_{1}x^{2}+a_{2}x-a_{3}$. But substituting $ a_{1}\mapsto-a_{1}$ should repair that.

Suppose that $ r_2\not\in\mathbb{Q}.$ Let $ r=r_{1}-r_{2}\in\mathbb{Q}.$ Case: $ r\neq0$ $ \Rightarrow r_{2}$ is a root of the polynomial: $ Q(x)=p(x+r)-p(x)=3rx^{2}+\left(3r^{2}+2a_{1}r\right)x+\left(r^{3}+a_{1}r^{2}+a_{2}r\right).$ Note that $ Q(x)$ is a quadratic polynomial with rational coefficients, hence $ r_{2}=t+v$ for some $ t\in\mathbb{Q}$ and $ v$ such that $ v^{2}\in\mathbb{Q}.$ From $ r_{2}\not\in\mathbb{Q},$ we have $ v\not\in\mathbb{Q},$ therefore $ t-v$ is a root of $ p(x).$ $ \Rightarrow\{r_{1},r_{2},r_{3}\}=\{t+v,t-v,-a_{1}-(t+v)-(t-v)\}=\{t+v,t-v,-a_{1}-2t\}$ $ \Rightarrow r_{1}-r_{2}\not\in\mathbb{Q},$ contradiction. Case: $ r=0$ $ \Rightarrow p(x)$ has multiple root $ r_{2}=r_{1}$ $ \Rightarrow r_{2}$ is a root of the polynomial: $ R(x)=\frac{d\left(p(x)\right)}{dx},$ which $ R(x)$ is a quadratic polynomial with rational coefficients, similarly to the first case, we get contradiction. Therefore $ r_{2}\in\mathbb{Q},$ and $ r_{1}=r_{2}+r\in\mathbb{Q},$ $ r_{3}=-a_{1}-r_{1}-r_{2}\in\mathbb{Q}.$

wya wrote: therefore $ t-v$ is a root of $ p(x).$ Can somebody tell me why we have the above remark?