# $\frac1a+\frac1b+\frac1c=0 \implies a^2+b^2+c^2=(a+b+c)^2$? [closed]

How to prove that $a^2+b^2+c^2=(a+b+c)^2$ given that $\frac1a+\frac1b+\frac1c=0$?

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## closed as off-topic by Davide Giraudo, dfeuer, user1337, Adriano, azimutAug 28 '13 at 22:00

This question appears to be off-topic. The users who voted to close gave this specific reason:

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It will be better if you show at least some of your work or whatever you have attempted ... that way people would be more likely to answer your questions –  user67258 Aug 3 '13 at 2:41

Expand the right side:

$$(a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc.$$

From the supplementary condition we have

$$\frac{1}{a}+\frac{1}{b} + \frac{1}{c} = 0$$

Or

$$\frac{ab+ac+bc}{abc} = 0.$$

Therefore $ab+ac+bc = 0$ and the result follows. (None of $a,b,c$ can be $0$ else their inverses would be undefined and so the supplementary condition would be ill-posed.)

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

since $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$.

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