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.
 A: Suppose $x^2 + 2x = q_1$ and $x^3 - 6x = q_2$ with $q_1, q_2\in \mathbb{Q}$. Clearly $[\mathbb{Q}(x):\mathbb{Q}] = 2$. Thus the polynomial $t^2 + 2t - q_1$ divides $t^3 - 6t - q_2$; that is, there exists some $\alpha$ such that
$$t^3 - 6t - q_2 = (t^2 +2t + q_1)(t - \alpha) = t^3 + (2 - \alpha)t^2 + (q_1 - 2\alpha)t - \alpha q_1.$$
Equating corresponding coefficients of $t^n$ gives $\alpha = 2$, $q_1 = -2,$ and $q_2 = 4$. Thus both polynomials are rational iff $x^2 + 2x + 2 = 0$; that is, iff $x = -1 \pm \sqrt{3}$.
A: I add yet another solution hopefully a simple one.
Since $x^2+2x$ is rational so too is $x^2+2x+1=(x+1)^2$. 
Let $y=x+1$, so $y^2$ is rational,  then $x^3-6x=y(y^2-3)-3y^2+5$ is rational, which means that $y(y^2-3)$ is rational. 
Now since $y$ is irrational and $y^2-3$ rational we must have $y^2-3=0$, thus $y=\pm \sqrt{3}$ and $x=-1\pm\sqrt{3}$
A: $q,r\in\Bbb Q,\,\ x^2 = \color{#0a0}{q\!-\!2x},\,\  r = x(x^2\!-\!6) = x(\color{#0a0}{q\!-\!2x}\!-6) = (q\!-\!6)x - 2(\color{#0a0}{q\!-\!2x}) = (\color{#c00}{q\!-\!2})x - 2q$  
Thus $\, x\not\in\Bbb Q\,\Rightarrow\, \color{#c00}{q=2}\,\Rightarrow\,r=-2q = -4,$ so the polynomials are $\,f = x^2\!+2x-2,\,$ and $x^3-6x+4 = (x\!-\!2)f,\,$ with irrational roots being the roots of $\,f,\,$ i.e.   $\,-1\pm \sqrt 3$
A: Suppose $q=x^2+2x$ is rational.
Then, by the quadratic formula, $x=\frac{-2 \pm \sqrt{4+4q}}{2}=-1 \pm \sqrt{\alpha}$, where $\alpha=q+1$ can be any rational number that is not a perfect square (since $x$ is irrational).
Now,
$$x^3-6x=(-1 \pm \sqrt{\alpha})^3-6(-1\pm\sqrt{\alpha}) = \pm \alpha^{3/2} - 3\alpha \mp 3 \sqrt{\alpha}+5$$
will be rational if and only if $\alpha^{3/2}-3\sqrt{\alpha}$ is (since the other terms are always rational). But this is equal to $\sqrt{\alpha}(\alpha-3)$. Since $\alpha-3$ is rational and $\sqrt{\alpha}$ irrational, the entire quantity will be rational iff it is zero: that is, iff $\alpha=3$.
Thus the numbers $x$ that satisfy the condition are $-1 \pm \sqrt{3}$.
A: 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.
