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: Well, assuming that $a,b,c\neq 0$, we can see that:
\begin{align}
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}&=\frac{1}{a+b+c}
\\ \\
\frac{ab+bc+ac}{abc}&=\frac{1}{a+b+c}
\\ \\
a^2b+abc+a^2c+ab^2+b^2c+abc+abc+bc^2+ac^2&=abc
\\ \\
(ab+b^2+ac+bc)c+(ab+b^2+ac+bc)a&=0
\\ \\
(ab+b^2+ac+bc)(c+a)&=0
\\ \\
(a+b)(b+c)(c+a)&=0
\end{align}
Now we know that $(a+b)(b+c)(c+a)=0$
$\therefore a=-b\text{ or }b=-c\text{ or }a=-c$.
We can use the fact $a=-b$ to our LHS:
$$\frac{1}{(-b)^{2013}}+\frac{1}{b^{2013}}+\frac{1}{c^{2013}}=-\frac{1}{b^{2013}}+\frac{1}{b^{2013}}+\frac{1}{c^{2013}}=\frac{1}{c^{2013}}$$
$$\text{RHS: }\frac{1}{-b^{2013}+b^{2013}+c^{2013}}=\frac{1}{c^{2013}}=LHS$$
A: 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 $$\pm\frac{1}{p}, p$$
