<s>[INST] Number of reals such that $\frac{1}{x-f(1)}+\frac{2}{x-f(2)}+\frac{3}{x-f(3)}=0$ 
If $f(x)= x^3+3x^2+6x+2009$ and $g(x)=\dfrac{1}{x-f(1)}+\dfrac{2}{x-f(2)}+\dfrac{3}{x-f(3)}.$
Then number of real solutions of $g(x)=0$ is

$\bf{My\; Try::}$ Given $f(x) = x^3+3x^2+6x+2009\;,$ Now
$$f'(x) = 3x^2+6x+6 = 3[(x+1)^2+1]>0\forall x\in \mathbb{R}$$
So $f(x)$ is an Strictly Increasing function.
So $f(1)<f(2)<f(3)\;,$ Now Let $f(1) = a$ and $f(2) = b$ and $f(3) = c\;,$
Then we get $a<b<c$
Now $$\displaystyle g(x) = \frac{1}{x-a}+\frac{2}{x-b}+\frac{3}{x-c}.$$
$$\displaystyle  = \frac{(x-b)\cdot (x-c)+2(x-c)\cdot (x-a)+3\cdot (x-a)\cdot (x-b)}{(x-a)\cdot (x-b)\cdot (x-c)}$$
Now How can i solve after that, help me
Thanks
 [/INST] Two solutions. It is enough to see that
$$
g^\prime(x)=-\sum_{i=1}^3 \frac{i}{(x-f(i))^2}<0.
$$
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