6. Real numbers $a, b, c$ satisfy $a+b+c=0$, and $a b c>0$, then the value of $a b+b c+c a$ is ( ).
(A) Positive number
(B) Negative number
(C) Zero
(D) A number whose sign cannot be determined


Answer: (B)
Solution: Since
$$
a+b+c=0, a b c>0,
$$

Therefore, $a, b, c$ are all non-zero,

Thus
$$
\begin{aligned}
a b+b c+c a & =\frac{1}{2}\left[(a+b+c)^{2}-a^{2}-b^{2}-c^{2}\right], \\
& =-\frac{1}{2}\left(a^{2}+b^{2}+c^{2}\right) \\
& 0$ indicates that among $a, b, c$, there must be two negatives and one positive. Without loss of generality, assume $a<0, b<0, c>0$, because
$$
\begin{array}{c}
a+c=-b>0, \\
c>-a>0,
\end{array}
$$

Therefore

Hence $|a|<|c|$, so
$$
-\frac{1}{a}>\frac{1}{c}
$$

That is
$$
\frac{1}{a}+\frac{1}{c}<0
$$

And
$$
\frac{1}{b}<0,
$$

So
$$
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}<0
$$

Thus
$$
a b+b c+c a=a b c \cdot\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)<0.
$$

Therefore, the answer is (B).