$1 \cdot 27$ If $a, b, c$ are non-zero real numbers and satisfy

and
$$
\begin{array}{l}
\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a}, \\
x=\frac{(a+b)(b+c)(c+a)}{a b c}, \quad \text { and } x<0,
\end{array}
$$

then $x$ equals
(A) -1 .
(B) -2 .
(C) -4 .
(D) -6 .
(E) -8 .
(29th American High School Mathematics Examination, 1978)
［Solution］Notice that

that is
$$
\begin{array}{c}
\frac{a+b-c}{c}+2=\frac{a-b+c}{b}+2=\frac{-a+b+c}{a}+2, \\
\frac{a+b+c}{c}=\frac{a+b+c}{b}=\frac{a+b+c}{a} .
\end{array}
$$

If $a+b+c=0$, these equations are satisfied.
If $a+b+c \neq 0$, then divide the latter equations by $a+b+c$, yielding
$$
\frac{1}{c}=\frac{1}{b}=\frac{1}{a}
$$

which means
$$
a=b=c .
$$

Thus, if $a+b+c=0$, then $\frac{(-c)(-a)(-b)}{a b c}=-1$; if $a=b=c$, then $x=\frac{8 a^{3}}{a^{3}}=8$, which contradicts the given condition $x<0$. Therefore, the answer is $(A)$.