$\begin{array}{l} \left\{ \begin{array}{l} \left( {x - y} \right)\left( {{x^2} + xy + {y^2} + 3} \right) = 3\left( {{x^2} + {y^2}} \right) + 2\left( 1 \right)\\ {x^2}y + {x^2} - 2x - 12 = 0\left( 2 \right) \end{array} \right.\\ \left( 1 \right) \Leftrightarrow \left( {x - y} \right)\left( {{x^2} + xy + {y^2}} \right) + 3\left( {x - y} \right) = 3{x^2} + 3{y^2} + 2\\ \Leftrightarrow {x^3} - {y^3} + 3x - 3y = 3{x^2} + 3{y^2} + 2\\ \Leftrightarrow {x^3} - 3{x^2} + 3x - 1 = {y^3} + 3{y^2} + 3y + 1\\ \Leftrightarrow {\left( {x - 1} \right)^3} = {\left( {y + 1} \right)^3}\\ \Leftrightarrow x - 1 = y + 1 \Leftrightarrow y = x - 2\\ \left( 2 \right) \to \left( 1 \right)\\ \Rightarrow {x^2}\left( {x - 2} \right) + {x^2} - 2x - 12 = 0\\ \Leftrightarrow {x^3} - {x^2} - 2x - 12 = 0\\ \Leftrightarrow \left( {x - 3} \right)\left( {{x^2} + 2x + 4} \right) = 0\\ \Leftrightarrow x = 3\\ \Rightarrow y = x - 2 = 1\\ \Rightarrow \left( {x;y} \right) = \left( {3;1} \right) \end{array}$
Vậy phương trình có nghiệm:
$\left\{ \begin{array}{l} x = 3\\ y = 1 \end{array} \right.$
