\(\left\{ \begin{array}{l}\left( {x + y} \right)\left( {{x^2} + {y^2}} \right) = 5\\\left( {x - y} \right)\left( {{x^2} - {y^2}} \right) = 3\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{x^3} + {y^3} + xy\left( {x + y} \right) = 5\\{x^3} + {y^3} - xy\left( {x + y} \right) = 3\end{array} \right.\)
Đặt \(\left\{ \begin{array}{l}{x^3} + {y^3} = a\\xy\left( {x + y} \right) = b\end{array} \right.\) ta có:
\(\left\{ \begin{array}{l}a + b = 5\\a - b = 3\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}2a = 8\\a - b = 3\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = 4\\b = 1\end{array} \right.\)
Suy ra
\(\begin{array}{l}\left\{ \begin{array}{l}{x^3} + {y^3} = 4\\xy\left( {x + y} \right) = 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{x^3} + {y^3} = 4\\3xy\left( {x + y} \right) = 3\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}xy\left( {x + y} \right) = 1\\{x^3} + {y^3} + 3xy\left( {x + y} \right) = 7\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}xy\left( {x + y} \right) = 1\\{\left( {x + y} \right)^3} = 7\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}xy\left( {x + y} \right) = 1\\x + y = \sqrt[3]{7}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}xy = \dfrac{1}{{\sqrt[3]{7}}}\\x + y = \sqrt[3]{7}\end{array} \right.\end{array}\)
\( \Rightarrow x,y\) là nghiệm của phương trình \({X^2} - \sqrt[3]{7}X + \dfrac{1}{{\sqrt[3]{7}}} = 0 \Leftrightarrow \left[ \begin{array}{l}X = ...\\X = ...\end{array} \right.\)
