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

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.