\(\Leftrightarrow\left\{{}\begin{matrix}x^3-xy^2-x^2y+y^3=3\\x^3+xy^2+x^2y+y^3=15\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x^3+y^3=9\\xy^2+x^2y=6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^3+y^3=9\\xy\left(x+y\right)=6\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x^3+y^3=9\\3xy\left(x+y\right)=18\end{matrix}\right.\) \(\Rightarrow\left(x+y\right)^3=27\)
\(\Rightarrow x+y=3\Rightarrow xy=\dfrac{6}{x+y}=2\) \(\Rightarrow\left\{{}\begin{matrix}x+y=3\\xy=2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y=3-x\\xy-2=0\end{matrix}\right.\)
\(\Rightarrow x\left(3-x\right)-2=0\Leftrightarrow-x^2+3x-2=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\Rightarrow y=2\\x=2\Rightarrow y=1\end{matrix}\right.\)
Vậy hệ có 2 cặp nghiệm \(\left(x;y\right)=\left(1;2\right);\left(2;1\right)\)
