<s>[INST] 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.
 [/INST] 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 $d$.
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