<s>[INST] $x$ is an irrational number such that $x^2 - 2x$ and $x^3 -5x$ are rational. 
If $x$ is an irrational number such that $x^2 - 2x$ and $x^3 -5x$ are rational numbers. What does $x^3 - 5x$ equals to?

 [/INST] From the fact that $x^2-2x$ is rational, with $x$ irrational, we conclude that $x$ has degree $2$ over $\Bbb Q$, so we can write it in the form $a+b\sqrt{D}$ for some rational $a$ and $b$ and squarefree $D\in\Bbb{Z}$.
Now we can write: $$x^2-2x = (a^2-2a+Db^2)+2b(a-1)\sqrt{D}\in\Bbb Q$$
Since $b\ne 0$, we can conclude that $a=1$, so $x=1+b\sqrt{D}$. Another calculation: $$x^3-5x = (3b^2D-4) + b(b^2D-2)\sqrt{D}\in\Bbb Q$$
We conclude that $b^2D=2$, so: $$x^3-5x=3b^2D-4=6-4=2$$
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