One, (Total 20 points) $m$ and $n$ are real numbers, and $m^{3}+$ $n^{3}+3 m n=1$. Find the value of $m+n$.


$$
\begin{array}{l}
\left\{\left(m^{3}+n^{3}\right)-\left(m^{2}-m n+n^{2}\right)\right\}+(m+n)^{2}-1=0, \\
\text { i.e., }\left(m^{2}-m n+n^{2}\right)(m+n-1)+(m+n+1)(m+n-1) \\
=0, \\
\quad(m+n-1)\left(m^{2}-m n+n^{2}+m+n+1\right)=0, \\
\frac{1}{2}(m+n-1)\left[(m-n)^{2}+(m+1)^{2}+(n+1)^{2}\right]=0 .
\end{array}
$$

When $m+n-1=0$, $m+n=1$;
When $(m-n)^{2}+(m+1)^{2}+(n+1)^{2}=0$, we have $m=n=-1$. Thus, $m+n=-2$.
Therefore, $m+n$ equals 1 or -2.