# 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.

• Option $c$ looks good Aug 17 '16 at 18:05
• Are you sure this is an Olympiad — the options? Aug 17 '16 at 18:05
• @QuestionAsker well, this is not an olympiad question but a mock test question Aug 17 '16 at 18:06
• Perhaps I am missing something, but surely the value of $\frac{1}{a^5} + \frac{1}{b^5} + \frac{1}{c^5}$ would always be $c$, since $c$ is the same - $\frac{1}{a^5} + \frac{1}{b^5} + \frac{1}{c^5}$? Aug 17 '16 at 18:09
• Oh! I'm really sorry it was my fault, I wrote the wrong option by mistake. Option c is actually what I'm going to edit now. I'm really really sorry :( Aug 17 '16 at 18:12

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 $d$.

• But, is there any way to prove it purely using algebra and not putting values and not doing hit-and-trial method. My argument for this is that, you are not provided with a statement like $a, b, c$ can be any integer. Rather, their values are probably well defined 'cause this statement is not in general case but a special case. This was what I was trying to do. Using pure algebra. Aug 17 '16 at 18:33
• I'm sorry, but I don't see how $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{abc}\overbrace{\iff}^{\text{multiply by a+b+c}} 3+2(ab+bc+ac)=1$. Is it $a + b + c$ before the $\iff$? Even so, how does the multiplication work? Aug 17 '16 at 18:36
• multiply both sides of the equation by $a+b+c$, the right side becomes $1$ and the left side becomes $1+ab+ac+ba+1+bc+ca+cb+1$ Aug 17 '16 at 18:49
• The subset of $\mathbb R^3$ satisfying $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ is a one sheeted paraboloid. and the function defined as $\frac{1}{a^5}+\frac{1}{b^5}+\frac{1}{c^5}$ restricted to the paraboloid takes on a ton of values. Aug 17 '16 at 18:52

$\frac 1a + \frac 1b + \frac 1c = \frac 1{a+b+c}\\ \frac {ab +ac + bc}{abc}= \frac 1{a+b+c}\\ (a+b+c)(ab +ac + bc) = abc\\ a^2b + a^2c + ab^2 +ac^2 + b^2c + bc^2 + 2abc = 0\\ (a^2b + ab^2) + c(a^2 + b^2 + 2ab + cb + ca)= 0\\ ab (a+b) + c((a+b)^2 + c(a+b))= 0\\ (a+b)(ab + c(a+b) + c^2) = 0\\ (a+b)(a(b + c) + c (b+c)) = 0\\ (a+b)(b+c)(a + c) = 0$

$a+b = 0$ 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

• From your result, I made an inference, please correct me if I'm wrong. That $\frac {1}{a^x} + \frac {1}{b^x} + \frac {1}{c^x} = \frac {1}{a^x + b^x + c^x}$ if such a condition like the equation one is given. Aug 18 '16 at 4:51
• Not true when $x$ is even. Aug 18 '16 at 21:08