\(\left\{{}\begin{matrix}\left(x+y\right)^3-3xy\left(x+y\right)=8\\x+y+2xy=2\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+y=a\\xy=b\end{matrix}\right.\) với \(a^2\ge4b\)
\(\Rightarrow\left\{{}\begin{matrix}a^3-3ab=8\\a+2b=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a^3-3ab=8\\a=2-2b\end{matrix}\right.\)
\(\Rightarrow\left(2-2b\right)^3-3b\left(2-2b\right)-8=0\)
\(\Leftrightarrow2b\left(4b^2-15b+15\right)=0\)
\(\Leftrightarrow b=0\Rightarrow a=2\Rightarrow\left(x;y\right)=\left(0;2\right);\left(2;0\right)\)