1. Given real numbers $a, b, c$ satisfy $a+b+c=0, a b c=8$, then the value of $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$.
(A) is positive;
(B) is zero;
(C) is negative;
(D) cannot determine whether it is positive or negative.
1. (C)

Since $a b c=8$, then $a, b, c$ are all non-zero.
Also, since $a+b+c=0$, we have
$$
\begin{array}{c}
(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2(a b+b c+c a)=0 . \\
Since \ a^{2}+b^{2}+c^{2}>0, \quad then \quad a b+b c+c a<0 .
\end{array}
$$

Therefore, $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{a b+b c+c a}{a b c}=\frac{a b+b c+c a}{8}<0$.