Polynomial and distinct roots Find all real m such that $x^3-2x^2-2x+m$ has 3 distinct rational roots.
Source: School exam paper. No idea why it seems so hard
I don't think rational root theorem works, since m is not necessarily integer (clearly it is rational though)
Assuming $m=\frac{r}{s}$, we need to check $sx^3-2sx^2-2sx+r=0$ for all $s,r$ coprime...

Let $a, b, c$ be the rational roots. Then we have 
$$a+b+c=2,\quad ab+bc+ca=-2, \quad m=-abc.$$
Replace $c=2-a-b$,
$$a^2+b^2+ab-2(a+b)-2=0.$$
(If $a=b$, then $a$ is irrational) Thus,
$$\Delta_b=(a-2)^2-4(a^2-2a-2)=-3a^2+4a+12$$
is a square rational.
The rest is easy.
Added after diner. Since 
$$-3a^2+4a+12=-3(a+\frac{2}{3})^2+4\times \frac{10}{3}$$ 
is a square. Let $\frac{3}{2}(a+\frac{2}{3})=\frac{p}{q}$ (suppose $\gcd(p,q)=1$), then 
$$-3p^2+30 q^2$$ 
is a square of integer, say $3r$, thus
$$-p^2+10q^2=3r^2.$$
It is clear that $2|p-r$. If both are even, then $q$ is even too, contradicting $\gcd(p,q)=1$. So $p$ and $r$ both are odd, then $p^2+3r^2=4 \mod(8)$, therefore $q$ is even, hence $10q^2=0 \mod (8)$, this is an contradiction with $p^2+3r^2=10 q^2$. 
If I didn't make mistake,  Conclusion: there is no $a, b, c$, there is  no $m$.
