$x$ 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?

 A: Note that $(x-1)^2=y$ is a rational number. So $x = 1 \pm \sqrt{y}$. Also,
$x^3-5x=3y-4 \pm\sqrt{y}(y-2)$ is rational, which is possible only when
$$y=2.$$
Therefore, $x = 1\pm \sqrt{2}$ and $x^3-5x = 3y-4 = 2$.
A: 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 $r= 1$:
$$ x^3-5x = (x^2-2x)q = qr = 2$$
A: The answer is $2$. (And I have to write at least 30 characters, so here they are...)
Added: I don't want to give away the whole answer, but here is an elementary way to do it: Write $x^2 - 2x = p/q$, solve with the quadratic equation, plug this into $x^3-5x$, observe that the square root term has to vanish, so the factor in front of it has to be zero, so you get an equation for $x$ (with two possible values, but both give the same answer for $x^3-5x$.)
A: By hypothesis there exist $a$ and $b$ such that $x$ is a root of the polynomials $f(X)=X^2-2X+a$ and $g(x)=X^3-5X+b$. Since $x$ is irrational, $[\mathbb Q(x):\mathbb Q]=2$ and the minimal polynomial of $x$ is $f(X)$. This implies that $g(X)$ is a multiple of $f(X)$, so
$$g(X)=(X-c)f(X)$$
and equalising the coefficients we find that $a=-1$, $b=-2$ and $c=-2$. Hence $x$ is a root of $X^2-2X-1$, and $x^3-5x=2$.
A: 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
$$
x^3 - 5 x = -1 + 3 \frac{m}{n} = 2.
$$
A: 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: $$x^3-5x=3b^2D-4=6-4=2$$
