1-40 Let $\frac{x}{x^{2}-m x+1}=1$, then the value of $\frac{x^{3}}{x^{6}-m^{3} x^{3}+1}$ is
(A)1.
(B) $\frac{1}{m^{3}+3}$.
(C) $\frac{1}{3 m^{2}-2}$.
(D) $\frac{1}{3 m^{2}+1}$.
(China Sichuan Province Junior High School Mathematics Competition, 1994)
[Solution] From the given, we know $x \neq 0$, so $\frac{x^{2}-m x+1}{x}=1$, which means $x+\frac{1}{x}=m+1$.

Thus, $\quad \frac{x^{6}-m^{3} x^{3}+1}{x^{3}}=x^{3}+\frac{1}{x^{3}}-m^{3}$
$$
\begin{array}{l}
=\left(x+\frac{1}{x}\right)^{3}-3\left(x+\frac{1}{x}\right)-m^{3} \\
=3 m^{2}-2,
\end{array}
$$

Therefore, $\frac{x^{3}}{x^{6}-m^{3} x^{3}+1}=\frac{1}{3 m^{2}-2}$. Hence, the answer is $(C)$.