$\frac{1}{x-a} + \frac{1}{x-b} + \frac{1}{x-c} = 0 $ has precisely two real roots Prove that given $ a < b < c $ this equation:  $$\frac{1}{x-a} + \frac{1}{x-b} + \frac{1}{x-c} = 0 $$ 
has precisely 2 real roots. 
I understand there are 3 point of discontinuities, but I have no idea how to prove this. Can you give me a hint? 
Thanks in advance.

Set $y=x-c$, so that $x=y+c$ and $x-a=y+c-a$, $x-b=y+c-b$. Set $r=c-a$ and $s=c-b$. Then your equation becomes
$$
\frac{1}{y+r}+\frac{1}{y+s}+\frac{1}{y}=0
$$
or
$$
y(y+s)+y(y+r)+(y+r)(y+s)=0.
$$
Simplifying you get
$$
3y^2+2(r+s)y+rs=0
$$
and the reduced discriminant is
$$
(r+s)^2-3rs=r^2-rs+s^2>0
$$
by well known facts about second degree polynomials.
Now $r>0$ and $s>0$, so $0$ is not a root. If we substitute $-r$ to $y$ we get
$$
3r^2-2r^2-rs+rs=r^2\ne0
$$
and similarly $-s$ is not a root. Thus both roots of the polynomial in $y$ are solutions of the equation in $y$ and therefore the original equation has two real solutions.
