II. (30 points) Given positive numbers $m, n$ are the roots of the quadratic equation $x^{2}+$ $p x+q=0$, and $m^{2}+n^{2}=3, m n=1$. Find the value of the polynomial $x^{3}-(\sqrt{5}-1) x^{2}-(\sqrt{5}-1) x+1994$.
$$
=, \because m, n>0, \therefore m+n=5 .
$$

Thus, $p=-(m+n)=-\sqrt{5}, q=mn=1$,
$$
\therefore x^{2}-\sqrt{5} x+1=0 \text {. }
$$
$$
\text { Also } \begin{array}{l} 
x^{3}-(\sqrt{5}-1) x^{2}-(\sqrt{5}-1) x+1994 \\
=\left(x^{3}+1\right)-(\sqrt{5}-1) x(x+1)+1993 \\
=(x+1)\left[\left(x^{2}-x+1\right)-(\sqrt{5}-1) x\right] \\
\quad+1993 \\
=(x+1)\left(x^{2}-\sqrt{5} x+1\right)+1993 \\
= 1993 .
\end{array}
$$