ข้อ 1
$\dfrac{a-b}{a+b}=x,\dfrac{b-c}{b+c}=y,\dfrac{c-a}{c+a}=z$
$x+1=\dfrac{2a}{a+b},1-x=\dfrac{2b}{a+b}$
$y+1=\dfrac{2b}{b+c},1-y=\dfrac{2c}{a+b}$
$z+1=\dfrac{2c}{c+a},1-c=\dfrac{2a}{c+a}$
$(x+1)(y+1)(z+1)=(1-z)(1-y)(1-x)$
$2xyz=-2(x+y+z)$
$xyz=-2553$ $(x+y+z=2553)$
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