$x$ 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?


Let $x^2 - 2 x = \frac{m}{n}$ for $m\in \mathbb{Z}, n\in \mathbb{N}$. We can solve this quadratic equation and obtain
$$
x = 1 \pm \sqrt{1 + \frac{m}{n}}
$$
Than we can substitute this $x$ this to $x^3 - 5x$
$$
x^3 - 5x = 1 \pm 3\sqrt{1 + \frac{m}{n}} + 3 \bigg(1 + \frac{m}{n}\bigg) \pm \sqrt{1 + \frac{m}{n}} \bigg(1 + \frac{m}{n}\bigg) - 
$$
$$
- 5\bigg(1 \pm \sqrt{1 + \frac{m}{n}}\bigg) =
$$
$$
= -1 + \frac{3m}{n} + \sqrt{1 + \frac{m}{n}} \bigg(\pm 3 \pm 1 \pm \frac{m}{n} \mp 5\bigg)
$$
Since this number should be rational we must have either
$$
3+1+\frac{m}{n} - 5 = -1 + \frac{m}{n} = 0
$$
or
$$
-3-1-\frac{m}{n} + 5 = 1- \frac{m}{n} = 0.
$$
Both cases give us $\frac{m}{n} = 1$ and
$$
x^3 - 5 x = -1 + 3 \frac{m}{n} = 2.
$$
