1. Given that $x, y, z, a, b$ are all non-zero real numbers, and satisfy $\frac{x y}{x+y}=\frac{1}{a^{3}-b^{3}}, \frac{y z}{y+z}=\frac{1}{a^{3}}, \frac{z x}{z+x}=\frac{1}{a^{3}+b^{3}}$, $\frac{x y z}{x y+y z+z x}=\frac{1}{12}$. Then the value of $a$ is ( ).
(A) 1
(B) 2
(C) -1
(D) -2
$$
-1 . B \text {. }
$$

Taking the reciprocal of each known equation, we get
$$
\begin{array}{l}
\frac{1}{x}+\frac{1}{y}=a^{3}-b^{3}, \text { (1) } \frac{1}{y}+\frac{1}{z}=a^{3}, \\
\frac{1}{z}+\frac{1}{x}=a^{3}+b^{3}, \text { (3) } \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=12 . \\
\frac{(1)+(2)+(3)}{2}, \text { we get } \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{3}{2} a^{3} .
\end{array}
$$
(1) $\frac{1}{y}+\frac{1}{z}=a^{3}$,
(3) $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=12$.
$\frac{(1)+(2)+(3)}{2}$, we get $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{3}{2} a^{3}$.
Substituting into (4) gives $\frac{3}{2} a^{3}=12$. Solving for $a$ yields $a=2$.