3
$\begingroup$

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.

$\endgroup$
5
  • 3
    $\begingroup$ Option $c$ looks good $\endgroup$
    – Asinomas
    Aug 17 '16 at 18:05
  • $\begingroup$ Are you sure this is an Olympiad — the options? $\endgroup$
    – Shuri2060
    Aug 17 '16 at 18:05
  • $\begingroup$ @QuestionAsker well, this is not an olympiad question but a mock test question $\endgroup$
    – Mayank M.
    Aug 17 '16 at 18:06
  • 1
    $\begingroup$ 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} $? $\endgroup$
    – Nico A
    Aug 17 '16 at 18:09
  • 1
    $\begingroup$ 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 :( $\endgroup$
    – Mayank M.
    Aug 17 '16 at 18:12
3
$\begingroup$

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 $d$.

$\endgroup$
4
  • $\begingroup$ 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. $\endgroup$
    – Mayank M.
    Aug 17 '16 at 18:33
  • $\begingroup$ 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? $\endgroup$ Aug 17 '16 at 18:36
  • $\begingroup$ multiply both sides of the equation by $a+b+c$, the right side becomes $1$ and the left side becomes $1+ab+ac+ba+1+bc+ca+cb+1$ $\endgroup$
    – Asinomas
    Aug 17 '16 at 18:49
  • $\begingroup$ 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. $\endgroup$
    – Asinomas
    Aug 17 '16 at 18:52
1
$\begingroup$

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

$a+b = 0$ 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

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$
    – Mayank M.
    Aug 18 '16 at 4:51
  • 1
    $\begingroup$ Not true when $x$ is even. $\endgroup$
    – Doug M
    Aug 18 '16 at 21:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.