\(...\Leftrightarrow\left(x-y\right)^3-1+3xy\left(x-y\right)-3xy=0\)
\(\Leftrightarrow\left(x-y-1\right)\left(x^2-2xy+y^2+x-y+1\right)+3xy\left(x-y-1\right)=0\)
\(\Leftrightarrow\left(x-y-1\right)\left(x^2-2xy+y^2+x-y+1+3xy\right)=0\)
\(\Leftrightarrow\left(x-y-1\right)\left(x^2+xy+y^2+x-y+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-y-1=0\left(1\right)\\x^2+xy+y^2+x-y+1=0\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow x=y+1\left(x;y\in Z\right)\)
\(\left(2\right)\Leftrightarrow2\left(x^2+xy+y^2+x-y+1\right)=0\)
\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(x^2+2x+1\right)+\left(y^2-2y+1\right)=0\)
\(\Leftrightarrow\left(x+y\right)^2+\left(x+1\right)^2+\left(y-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\x+1=0\\y-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=y+1\left(x;y\in Z\right)\\\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\end{matrix}\right.\) thỏa mãn yêu cầu đề bài
