$x$ 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?


Since $x\ne 0$ and $x\ne 2$ we have  $${x^3-5x\over x^2-2x} = {x^2-5\over x-2}= q\in \mathbb{Q}$$
So $$x^2-5 = xq-2q \;\;\;\;{\rm and }\;\;\;\;x^2 =2x+r\in \mathbb{Q}$$
so $$ 2x+r-5 = xq-2q\implies (q-2)x =-2q-r+5$$ so if $q\ne 2$ we get $$x={-2q-r+5\over q-2}\;\;\;$$
but this can't be since $x$ is irational and left side is rational so $q=2$ and $r= 1$:
$$ x^3-5x = (x^2-2x)q = qr = 2$$
