1. ## Opposite number proof

Hello, I am stuck with this problem from Gelfand's Algebra book:
Problem 126. Prove that if $\frac{1}{a+b+c}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c }$, then there are two opposite numbers among $a, b, c (i.e: a=-b, a=-c, b=-c)$
To be honest, I don't understand the question.

2. ## Re: Opposite number proof

Hello, DIOGYK!

I think I got it . . .

$\text{Prove that if }\,\frac{1}{a+b+c}\:=\:\frac{1}{a}+\frac{1}{b}+ \frac{1}{c}$

$\text{then there are two opposite numbers among }a, b, c.$

$\text{That is: }\,a=\text{-}b\,\text{ or }\,b=\text{-}c\,\text{ or }\,a=\text{-}c.$

We have: . $\frac{1}{a+b+c} \;=\;\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$

. . . . . . . . $\frac{1}{a+b+c} \;=\;\frac{bc + ac + ab}{abc}$

. . . . . . . . . . . . $abc \;=\;(a+b+c)(bc+ac+ab)$

. . . . . . . . . . . . $abc \;=\;abc + a^2c + a^2b + b^2c + abc + bc^2 + ac^2 + abc$

. . $a^2b + a^2c + ab^2 + 2abc + ac^2 + b^2c + bc^2 \;=\;0$

n . $a^2(b + c) + a(b^2 + 2bc + c^2) + bc(b+c) \;=\;0$

. . . . . . . $a^2(b+c) + a(b+c)^2 + bc(b+c) \;=\;0$

Factor: . . . . . $(b+c)\big[a^2 + a(b+c) + bc\big] \;=\;0$

. . . . . . . . . . . $(b+c)\big[a^2 + ab + ac + bc\big] \;=\;0$

. . . . . . . . . . $(b+c)\big[a(a+b) + c(a+b)\big] \;=\;0$

Factor: . . . . . . . . . $(a+b)(b+c)(a+c) \;=\;0$

We see that the product of three factors equals zero.
Hence, at least one of the factors must be zero.

Therefore: . $\begin{Bmatrix}a+b \:=\:0 & \Rightarrow & a \:=\:\text{-}b \\ & \text{or} \\ b+c \:=\:0 & \Rightarrow & b \:=\:\text{-}c \\ & \text{or} \\ a+c\:=\:0 & \Rightarrow & a \:=\:\text{-}c \end{Bmatrix}$

3. ## Re: Opposite number proof

Originally Posted by Soroban
Hello, DIOGYK!

I think I got it . . .

We have: . $\frac{1}{a+b+c} \;=\;\frac{1}{a} + \frac{1}{b} + \frac{1}{c}$

. . . . . . . . $\frac{1}{a+b+c} \;=\;\frac{bc + ac + ab}{abc}$

. . . . . . . . . . . . $abc \;=\;(a+b+c)(bc+ac+ab)$

. . . . . . . . . . . . $abc \;=\;abc + a^2c + a^2b + b^2c + abc + bc^2 + ac^2 + abc$

. . $a^2b + a^2c + ab^2 + 2abc + ac^2 + b^2c + bc^2 \;=\;0$

n . $a^2(b + c) + a(b^2 + 2bc + c^2) + bc(b+c) \;=\;0$

. . . . . . . $a^2(b+c) + a(b+c)^2 + bc(b+c) \;=\;0$

Factor: . . . . . $(b+c)\big[a^2 + a(b+c) + bc\big] \;=\;0$

. . . . . . . . . . . $(b+c)\big[a^2 + ab + ac + bc\big] \;=\;0$

. . . . . . . . . . $(b+c)\big[a(a+b) + c(a+b)\big] \;=\;0$

Factor: . . . . . . . . . $(a+b)(b+c)(a+c) \;=\;0$

We see that the product of three factors equals zero.
Hence, at least one of the factors must be zero.

Therefore: . $\begin{Bmatrix}a+b \:=\:0 & \Rightarrow & a \:=\:\text{-}b \\ & \text{or} \\ b+c \:=\:0 & \Rightarrow & b \:=\:\text{-}c \\ & \text{or} \\ a+c\:=\:0 & \Rightarrow & a \:=\:\text{-}c \end{Bmatrix}$
Sorry for the late reply and thank you for your help, I appreciate it. Your explanation was very clear. : )