ta có điều kiện \(x\ne0;y\ne0\)ta có
\(\frac{1}{x^3}+\frac{1}{y^3}+3=x^3y^3\Leftrightarrow\left(\frac{1}{x}\right)^3+\left(\frac{1}{y}\right)^3+\left(-x^3y^3\right)=3.\frac{1}{x}.\frac{1}{y}.\left(-xy\right)\Leftrightarrow\orbr{\begin{cases}\frac{1}{x}=\frac{1}{y}=-xy\\\frac{1}{x}+\frac{1}{y}-xy=0\end{cases}}\)
TH1 : ta có \(\frac{1}{x}=\frac{1}{y}=-xy\Leftrightarrow\hept{\begin{cases}x=y\\1=-x^2y\end{cases}\Leftrightarrow\hept{\begin{cases}x=-1\\y=-1\end{cases}}}\)(thử zô (1) ko thỏa mãn )
TH2 :ta có \(\frac{1}{x}+\frac{1}{y}-xy=0\Leftrightarrow x+y=\left(xy\right)^2\)ta có
\(\left(1\right)\Leftrightarrow\frac{1}{x^2}+\frac{1}{y^2}=3+x^2y^2\Leftrightarrow xy\left(3xy+2\right)=0\Leftrightarrow xy=-\frac{2}{3}\)
\(\hept{\begin{cases}\frac{1}{x^2}+\frac{1}{y^2}=3+x^2y^2\left(1\right)\\\frac{1}{x^3}+\frac{1}{y^3}+3=x^3y^3\left(2\right)\end{cases}\Leftrightarrow\hept{\begin{cases}\frac{1}{x^2}+\frac{1}{y^2}=3+x^2y^2\\\orbr{\begin{cases}\frac{1}{x}=\frac{1}{y}=-xy\\\frac{1}{x}+\frac{1}{y}-xy=0\end{cases}}\end{cases}}}\)zậy \(\hept{\begin{cases}x+y=\left(xy\right)^2\\xy=-\frac{2}{3}\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{2+\sqrt{58}}{9}\\y=\frac{2-\sqrt{58}}{9}\end{cases}hoặc\hept{\begin{cases}x=\frac{2-\sqrt{58}}{9}\\y=\frac{2+\sqrt{58}}{9}\end{cases}}}}\)
