Prove that $x\in\mathbb Q$ Let $a\in\mathbb Q$ and $a>\dfrac43$. Let $x\in\mathbb R$ and $x^2-ax,x^3-ax\in\mathbb Q$. Prove that $x\in\mathbb Q$.
EDIT:
Thsi is my attempt:
Let $x^2-ax=b$ and $x^3-ax=q$ for some $b,q\in\mathbb{Q}$. Then I tried to write $x^2$ and $x^3$ in linear terms. I got $x^2=ax+b$ and substituting in second equation I got
$$
x\cdot x^2-ax=q\\
x(ax+b)-ax=q\\
ax^2+bx-ax=q\\
a(ax+b)+bx-ax=q\\
a^2x+ab+bx-ax=q
$$
But, what next? Since it is known that $a>\dfrac43$ I know that I must use discriminant of some quadratic equation, but how to get quadratic equation from here?

Let $x^2-ax=c$ and $x^3-ax=d$. Let $P(y)$ be the minimal polynomial of $x$ over $Q$. 
Then $P(Y)| Y^2-aY-c$ and $P(y) | Y^3-aY-d$. In particular the degree of $P$ is at most two.
Case 1  $\deg P=1$, it follows immediately that $x \in \mathbb Q$. 
Case 2  $\deg P=2$, then $P(Y)=Y^2-aY-c$ and hence
$$ Y^2-aY-c| Y^3-aY-d \,.$$
This shows that
$$Y^3-aY-d = (Y^2-aY-c)(Y-e)$$
The coefficient of $Y^2$ on the RHS is $-(a+e)$ which shows that $e=a$. This implies
$$Y^3-aY-d = (Y^2-aY-c)(Y+a)=Y^3-Y(a^2+c)-ac$$
Therefore 
$$a^2+c=a$$
$$d=ac$$
The first relation implies
$$P(Y)=Y^2-ay+a^2-a$$
As $P$ has real solutions $\Delta \geq 0$ and hence $a^2-4(a^2-a)>0 \Rightarrow 4a-3a^2 >0$, which contradicts $a>\frac{4}{3}$.
Note For $a=\frac{1}{2}$ we get 
$$P(Y)=Y^2-\frac{1}{2}Y-\frac{1}{4}$$
and
$$c=\frac{1}{4}, d=\frac{1}{8} \,.$$
As
$$(Y^2-\frac{1}{2}Y-\frac{1}{4})(Y+\frac{1}{2}) = Y^3-\frac{1}{2}Y-\frac{1}{8}$$
it follows that the roots 
$$x_{1,2} =\frac{1 \pm \sqrt{5}}{4}$$
of 
$$Y^2-\frac{1}{2}Y-\frac{1}{4}=0 $$
satisfy
$$x^2-\frac{1}{2}x=\frac{1}{4}$$
$$x^3-\frac{1}{2}x=\frac{1}{8}$$
So the condition $a > \frac{4}{3}$ is indeed needed.
