smc

From the equation $$\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a},$$ we add $2$ to each fraction to get $$\frac{a+b+c}{c}=\frac{a+b+c}{b}=\frac{a+b+c}{a}.$$ We perform casework on $a+b+c:$ If $a+b+c\ne 0,$ then $a=b=c,$ from which $x=\frac{(2a)(2a)(2a)}{a^3}=8.$ However, this contradicts the precondition $x<0.$ If $a+b+c=0,$ then $x=\frac{(-c)(-a)(-b)}{abc}=\boxed{-1}.$