If $a,b,c$ are non-zero real numbers such that \[\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a},\] and \[x=\frac{(a+b)(b+c)(c+a)}{abc},\] and $x<0,$ then $x$ equals
$\textbf{(A) }{-}1\qquad \textbf{(B) }{-}2\qquad \textbf{(C) }{-}4\qquad \textbf{(D) }{-}6\qquad  \textbf{(E) }{-}8$
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 $a+b+c\neq0,$ 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 $x=\frac{(-c)(-a)(-b)}{abc}=\boxed{\textbf{(A) }{-}1}.$
~MRENTHUSIASM