Determine the irrational numbers $x$ such that both $x^2+2x$ and $x^3-6x$ are rational numbers I did not make any progress. The problem is from RMC 2008.
The only idea that I have is:
Try to find sets of irrational numbers such that every number in the set multiplied by another number in the set yields a rational number.

Let $x$ be an irrational number with
$$x^{2} + 2x = \frac{p_{1}}{q_{1}},\ x^{3} - 6x = \frac{p_{2}}{q_{2}}$$
for some integers $p_{1}, q_{1}, p_{2}, q_{2}$ such that $q_{1}, q_{2} \neq 0$ and $(p_{1}, q_{1}) = (p_{2}, q_{2}) = 1.$
Then we have
$$2x^{2} - (6 + \frac{p_{1}}{q_{1}})x + \frac{p_{2}}{q_{2}} = 0,$$
so that
$$x = \frac{6 + \dfrac{p_{1}}{q_{1}} \pm \sqrt{(6 + \dfrac{p_{1}}{q_{1}})^{2} - \dfrac{8p_{2}}{q_{2}}}}{4},$$
whence if 
$$(6 + \dfrac{p_{1}}{q_{1}})^{2} - \dfrac{8p_{2}}{q_{2}} > 0$$
and is not a perfect square,
then $x$ is irrational.
