<s>[INST] $\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.
 [/INST] HINT:
Rearrange the equation to get $3x^2-2(a+b+c)x+ab+bc+ca=0$
$$\text{The discriminant }\{2(a+b+c)\}^2-4\cdot3\cdot(ab+bc+ca)$$
$$=4(a^2+b^2+c^2-ab-bc-ca)$$
$$=4\cdot\frac{\{(a-b)^2+(b-c)^2+(c-a)^2\}}2>0 \text{  for distinct real }a,b,c$$
What can we conclude from this?
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