1.49 Given real numbers $a, b, c$ satisfy the equations $a+b+c=0$, and $abc=8$, then the value of $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$
(A) is positive.
(B) is zero.
(C) is negative.
(D) cannot be determined as positive or negative.
(China Junior High School Mathematics League, 1987)
[Solution] $\because a b c=8, \therefore a, b, c$ are all non-zero.
Also, $\because a+b+c=0$, then we have
\[
\begin{array}{l}
\quad(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(a b+a c+c a)=0 . \\
\because \quad a^{2}+b^{2}+c^{2}>0, \quad \therefore \quad a b+b c+c a<0 \\
\text { Therefore, } \quad \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{a b+b c+c a}{a b c}=\frac{a b+b c+c a}{8}<0 .
\end{array}
\]

Hence, the correct choice is $(C)$.