A confusing question related to three variables with fractions.. Recently, I had a mock-test of a Mathematics Olympiad. There was a question which not only I but my friends too were not able to solve. The question goes like this:  
If, 
$$ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{a+b+c} $$
Then what is the value of
$$ \frac{1}{a^5} + \frac{1}{b^5} + \frac{1}{c^5} $$  
To, solve this question, I used a variety of ways like:
1) transposing variables in the first equation and,
2) putting a whole power of five to both the sides of the equation one. But, I was unable to find the solution.  
The options were -- (a) 1 , (b) 0 , (c) $ \frac{1}{a^5 + b^5 + c^5} $ , (d) None of them.  
So, I require any possible help. And, a complete answer would be most welcome. Thanks in advance.
 A: Options $1$ and $2$ are wrong, take $a=1,b=-1,c=-1$.
Now notice given non-zero real number $a,b,c$ we have:
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{abc}\overbrace{\iff}^{\text{multiply by a+b+c}} 3+2(ab+bc+ac)=1\iff ab+bc+ac=-1$.
Analogously we have $\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{1}{a^5+b^5+c^5}\iff a^5b^5+b^5c^5+a^5c^5=-1$.
Taking $a=2,b=2,c=\frac{-5}{4}$ satisfies $ab+bc+ac=-1$ but not $a^5b^5+b^5c^5+a^5c^5=-1$.
So the answer is $d$.
A: $\frac 1a + \frac 1b + \frac 1c = \frac 1{a+b+c}\\
\frac {ab +ac + bc}{abc}= \frac 1{a+b+c}\\
(a+b+c)(ab +ac + bc) = abc\\
a^2b + a^2c + ab^2 +ac^2 + b^2c + bc^2 + 2abc = 0\\
(a^2b + ab^2) + c(a^2 + b^2 + 2ab + cb + ca)= 0\\
ab (a+b) + c((a+b)^2 + c(a+b))= 0\\
(a+b)(ab + c(a+b) + c^2) = 0\\
(a+b)(a(b + c) + c (b+c)) = 0\\
(a+b)(b+c)(a + c) = 0$
$a+b = 0$ or $a+c = 0$ or $b+c =0$
Suppose $a=-b$
then 
$\frac 1{a^5} + \frac 1{b^5} + \frac 1{c^5} = \frac 1{c^5}$
And going throught the other possibilities it becomes clear that c) is correct
