1.56 When $x=\frac{1+\sqrt{1994}}{2}$, the value of the polynomial $\left(4 x^{3}-1997 x-1994\right)^{20001}$ is
(A) 1.
(B) -1.
(C) $2^{2001}$.
(D) $-2^{2001}$.
(China Junior High School Mathematics League, 1994)
[Solution] $\because \quad x=\frac{1+\sqrt{1994}}{2}$,
$\therefore(2 x-1)^{2}=1994$, i.e., $4 x^{2}-4 x-1993=0$. Therefore,
$$
\begin{aligned}
& \left(4 x^{3}-1997 x-1994\right)^{2001} \\
= & {\left[\left(4 x^{2}-4 x-1993\right) x+\left(4 x^{2}-4 x-1993\right)-1\right]^{2001} } \\
= & (-1)^{2001}=-1 .
\end{aligned}
$$

Hence, the answer is $(B)$.