$\begin{array}{l}\text { 2. If } \frac{y}{x}+\frac{x}{z}=a, \frac{z}{y}+\frac{y}{x}=b, \frac{x}{z}+\frac{z}{y}=c, \\ \text { then }(b+c-a)(c+a-b)(a+b-c)=\end{array}$
2.8.

Notice that,
$$
\begin{array}{l}
b+c-a=\left(\frac{z}{y}+\frac{y}{x}\right)+\left(\frac{x}{z}+\frac{z}{y}\right)-\left(\frac{y}{x}+\frac{x}{z}\right) \\
=\frac{2 z}{y} .
\end{array}
$$

Similarly, $c+a-b=\frac{2 x}{z}, a+b-c=\frac{2 y}{x}$.
Therefore, $(b+c-a)(c+a-b)(a+b-c)$ $=\frac{2 z}{y} \cdot \frac{2 x}{z} \cdot \frac{2 y}{x}=8$.