Ta có: \(\left\{{}\begin{matrix}x^3+y^2=2\\x^2+y^3=2\end{matrix}\right.\) \(\Rightarrow x^3+y^2=x^2+y^3\Leftrightarrow x^3-x^2=y^3-y^2\Leftrightarrow x^2\left(x-1\right)=y^2\left(y-1\right)\)
\(\Rightarrow\) x=y
\(\Rightarrow\)\(x^3+y^2=2\Leftrightarrow x^3+x^2=2\Leftrightarrow x=1\)\(\Rightarrow y=1\)