1.6 For any positive real numbers $x, y, z$, the product $(x+y+z)^{-1}$ $\left(x^{-1}+y^{-1}+z^{-1}\right)(x y+y z+z x)^{-1}\left[(x y)^{-1}+(y z)^{-1}+(z x)^{-1}\right]$ equals
(A) $x^{-2} y^{-2} z^{-2}$.
(B) $x^{-2}+y^{-2}+z^{-2}$.
(C) $(x+y+z)^{-2}$.
(D) $\frac{1}{x y z}$.
(E) $\frac{1}{x y+y z+z x}$.
(32nd American High School Mathematics Examination, 1981)
【Sol】 Convert the negative exponents to fractions and then simplify as follows:
$$
\begin{aligned}
& (x+y+z)^{-1}\left(x^{-1}+y^{-1}+z^{-1}\right)(x y+y z+z x)^{-1} \\
& \cdot\left[(x y)^{-1}+(y z)^{-1}+(z x)^{-1}\right] \\
= & \frac{1}{x+y+z}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\left(\frac{1}{x y+y z+z x}\right)\left(\frac{1}{x y}+\frac{1}{y z}+\frac{1}{z x}\right) \\
= & \frac{1}{x+y+z}\left(\frac{x y+y z+z x}{x y z}\right)\left(\frac{1}{x y+y z+z x}\right)\left(\frac{x+y+z}{x y z}\right) \\
= & \frac{1}{(x y z)^{2}}=x^{-2} y^{-2} z^{-2} .
\end{aligned}
$$

Therefore, the answer is (A).