\begin{array}{l}-1 . x=y=1 \text {. } \\ \text { Let } x=1+\delta_{1}, y=1+\delta_{2}\left(\delta_{1} 、 \delta_{2} \geqslant-1\right) \text {. } \\ \text { Then } x^{3}+y^{3}-3 x y+1 \\ =\left(1+\delta_{1}\right)^{3}+\left(1+\delta_{2}\right)^{3}- \\ 3\left(1+\delta_{1}\right)\left(1+\delta_{2}\right)+1 \\ =\left(\delta_{1}+\delta_{2}+3\right)\left(\delta_{1}^{2}-\delta_{1} \delta_{2}+\delta_{2}^{2}\right)=0 \text {. } \\\end{array}
But \(\delta_{1}+\delta_{2}+3>0\), thus,
\begin{array}{l}\delta_{1}^{2}-\delta_{1} \delta_{2}+\delta_{2}^{2}=0 \Rightarrow \delta_{1}=\delta_{2}=0 \\ \Rightarrow x=y=1\end{array}
1. The non-negative real solutions to the equation $x^{3}+y^{3}-3 x y+1=0$ are $\qquad$ .