How to prove this equation? $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = \frac{1}{a+b+c}$ Suppose $a, b$, and $c$ are nonzero real numbers which satisfy the equation: $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = \frac{1}{a+b+c}$
Prove: if $n$ is an odd integer, then $a^n + b^n + c^n=(a+b+c)^n$
I was thinking of relating this to natural logs and derivatives somehow but then i'm not even sure how that would help.

Note that:
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = \frac{1}{a+b+c}$$
$$\frac{ab+ac+bc}{abc}=\frac{1}{a+b+c}$$
$$(ab+ac+bc)(a+b+c)-abc=0$$
$$(a+b)(b+c)(c+a)=0$$
then $a=-b$, $a=-c$ or $b=-c$. 
If $a=-b$, how $n$ is odd, then 
$$a^n + b^n + c^n=(-b)^n + b^n + c^n=c^n$$ and $$(a+b+c)^n=(-b+b+c)^n=c^n$$
you continue...
