\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^3-2xy\left(x+y\right)=32\\x^2y^2\left[\left(x+y\right)^2-2xy\right]=128\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+y=a\\xy=b\end{matrix}\right.\) với \(a^2\ge4b\)
\(\Rightarrow\left\{{}\begin{matrix}a\left(a^2-2b\right)=32\\b^2\left(a^2-2b\right)=128\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a^3-2ab=32\\\frac{b^2}{a}=4\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a^3-2ab=32\\a=\frac{b^2}{4}\end{matrix}\right.\)
\(\Rightarrow\frac{b^6}{64}-\frac{b^3}{2}=32\)
\(\Leftrightarrow\frac{1}{64}b^6-\frac{1}{2}b^3-32=0\Rightarrow\left[{}\begin{matrix}b^3=64\\b^3=-32\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}b=4\Rightarrow a=4\\b=-2\sqrt[3]{4}\Rightarrow a=2\sqrt[3]{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+y=4\\xy=4\end{matrix}\right.\\\left\{{}\begin{matrix}x+y=2\sqrt[3]{2}\\xy=-2\sqrt[3]{4}\end{matrix}\right.\end{matrix}\right.\) theo Viet đảo x và y là nghiệm:
\(\left[{}\begin{matrix}t^2-4t+4=0\\t^2-2\sqrt[3]{2}t-2\sqrt[3]{4}=0\end{matrix}\right.\) \(\Rightarrow t=...\)
