[tex]\it \dfrac{-(b+c)}{a(a+b+c)} = \dfrac{b+c}{bc} \Rightarrow 0 = \dfrac{b+c}{bc} +\dfrac{b+c}{a(a+b+c)} \Rightarrow
\\ \\ \\
\Rightarrow (b+c)\left(\dfrac{1}{bc} +\dfrac{1}{a^2+ab+ac}\right) =0 \Rightarrow
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\Rightarrow (b+c)\left(\dfrac{a^2+ab+ac+bc}{bc(a^2+ab+ac)}\right) =0 \Rightarrow
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(b+c)(a^2+ab+ac+bc) =0 \Rightarrow(b+c)[a(a+b)+c(a+b)]=0 \Rightarrow
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\Rightarrow (b+c)(a+b)(a+c) =0 [/tex]
[tex]\it \Rightarrow \begin{cases} \it b+c=0\Rightarrow b=-c\Rightarrow |b|=|c|
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\it a+b=0\Rightarrow a=-b\Rightarrow |a|=|b|
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\it a+c=0\Rightarrow a=-c\Rightarrow |a|=|c|
\\ \\ \end{cases}[/tex]
