$\begin{array}{l} \left\{ \begin{array}{l} {x^3} + 3x = {y^3} - 8\left( 1 \right)\\ {x^2} + {y^2} = y + 2\left( 2 \right) \end{array} \right.\\ \left( 1 \right) + 3.\left( 2 \right)\\ \Rightarrow {x^3} + 3{x^2} + 3x + 3{y^2} = {y^3} + 3y - 1 - 1\\ \Leftrightarrow \left( {{x^3} + 3{x^2} + 3x + 1} \right) = {y^3} - 3{y^2} + 3y - 1\\ \Leftrightarrow {\left( {x + 1} \right)^3} = {\left( {y - 1} \right)^3}\\ \Leftrightarrow x + 1 = y - 1\\ \Leftrightarrow x - y = - 2 \Rightarrow x = y - 2\left( 3 \right)\\ \left( 2 \right) \Rightarrow {x^2} + {y^2} = y + 2\\ \Leftrightarrow {\left( {y - 2} \right)^2} + {y^2} = y + 2\\ \Leftrightarrow 2{y^2} - 4y + 4 = y + 2\\ \Leftrightarrow 2{y^2} - 5y + 2 = 0\\ \Leftrightarrow \left( {y - 2} \right)\left( {2y - 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} y = 2 \Rightarrow x = 0\\ y = \dfrac{1}{2} \Rightarrow x = - \dfrac{3}{2} \end{array} \right.\\ \Rightarrow \left( {x;y} \right) = \left( {0;2} \right),\left( { - \dfrac{3}{2};\dfrac{1}{2}} \right)\\ \end{array}$
