6. If positive integers $a, b, c, x, y, z$ satisfy
$$
a x=b+c, b y=a+c, c z=a+b,
$$
then the number of possible values for the product $x y z$ is ( ).
(A) 2
(B) 3
(C) 4
(D) infinitely many
6. B.

If the three numbers $a, b, c$ are all equal, then $x=y=z=2$, at this time, $xyz=8$.

If the three numbers $a, b, c$ are not all equal, let's assume $c$ is the largest number, then $a+b<2c$. According to $z=\frac{a+b}{c}$ being a positive integer, we get $z=1, c=a+b$.
$$
\begin{array}{l}
\text { Hence } x=\frac{b+c}{a}=\frac{a+2b}{a}=1+\frac{2b}{a}, \\
y=\frac{a+c}{b}=\frac{2a+b}{b}=1+\frac{2a}{b} .
\end{array}
$$

Since $a, b, x, y$ are positive integers, then $\frac{2b}{a}, \frac{2a}{b}$ are also positive integers. Let $\frac{2b}{a}=m, \frac{2a}{b}=n(m \leqslant n)$.
From $mn=4$, we get $m=n=2$ or $m=1, n=4$.
Therefore, $(x, y, z)=(3,3,1)$ or $(2,5,1)$.
Thus, $xyz=9$ or $xyz=10$.
In fact, when $(a, b, c)=(1,1,1)$, $(x, y, z)=(2,2,2)$;
when $(a, b, c)=(1,1,2)$,
$(x, y, z)=(3,3,1)$;
when $(a, b, c)=(1,2,3)$,
$(x, y, z)=(5,2,1)$.
Therefore, the product $xyz$ can take the values $8, 9, 10$.