\(\Leftrightarrow x^3+y^3+xy\left(x+y\right)=x^3-y^3-3xy\left(x-y\right)\)
\(\Leftrightarrow2y^3+xy\left(4x-2y\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=0\\y+x\left(2x-y\right)=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow y+2x^2-xy=0\Leftrightarrow2x^2=y\left(x-1\right)\)
Với \(x=1\) ko phải nghiệm, với \(x\ne1\)
\(\Rightarrow y=\frac{2x^2}{x-1}=2x+2+\frac{2}{x-1}\)
\(\Rightarrow x-1=Ư\left(2\right)=\left\{-2;-1;1;2\right\}\)
\(\Rightarrow x=...;y=...\)