Using the hypothesis $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}$ to prove something else Assuming that $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}$$
Is it possible to use this fact to prove something like:
$$\frac{1}{a^{2013}}+\frac{1}{b^{2013}}+\frac{1}{c^{2013}}=\frac{1}{a^{2013}+b^{2013}+c^{2013}}$$
Just curious.  Thanks for the help!

A simpler way to see that one of $a = -b$ or $b = -c$ or $c = -a$ is true.
wlog, we can assume $abc = 1$ (why?).
So we can assume $a,b,c$ are roots of $x^3 - px^2 + qx -1 = 0$.
Your relation gives us that $q = \frac{1}{p}$ (Vieta's formulas)
Thus we get $a,b,c$ are roots of
$$x^2(x - p) + \frac{1}{p}(x - p) = 0$$
Thus the roots are $$\pm\frac{1}{p}, p$$
