Answer:
Take the first two expressions (you can actually take any two expressions): $\frac{a+b-c}{c}=\frac{a-b+c}{b}$.
$\frac{a+b}{c}=\frac{a+c}{b}$
$ab+b^2=ac+c^2$
$a(b-c)+b^2-c^2=0$
$(a+b+c)(b-c)=0$
$\Rightarrow a+b+c=0$ OR $b=c$
The first solution gives us $x=\frac{(-c)(-a)(-b)}{abc}=-1$.
The second solution gives us $a=b=c$, and $x=\frac{8a^3}{a^3}=8$, which is not negative, so this solution doesn't work.
Therefore, $x=-1\Rightarrow\boxed{A}$.
Step-by-step explanation:
