giúp em với ạ em cần gấp $\left \{ {8{x^{3} - 12x^{2} + 8x = y^{3} + y + 2} \atop {x^{2} + y^{2} = 2}} \right.$

$\begin{array}{l}
\left\{ \begin{array}{l}
8{x^3} - 12{x^2} + 8x = {y^3} + y + 2\left( 1 \right)\\
{x^2} + {y^2} = 2\left( 2 \right)
\end{array} \right.\\
\left( 1 \right) \Leftrightarrow 8{x^3} - 12{x^2} + 6x - 1 + 2x = {y^3} + y + 1\\
 \Leftrightarrow {\left( {2x - 1} \right)^3} + 2x = {y^3} + y + 1\\
 \Leftrightarrow {\left( {2x - 1} \right)^3} - {y^3} + 2x - y - 1 = 0\\
 \Leftrightarrow \left( {2x - y - 1} \right)\left[ {{{\left( {2x - 1} \right)}^2} + \left( {2x - 1} \right)y + {y^2}} \right] + \left( {2x - y - 1} \right) = 0\\
 \Leftrightarrow \left( {2x - y - 1} \right)\left[ {\underbrace {{{\left( {2x - 1} \right)}^2} + \left( {2x - 1} \right)y + {y^2} + 1}_{ > 0}} \right] = 0\\
 \Leftrightarrow 2x - y - 1 = 0 \Leftrightarrow y = 2x - 1\\
\left( 3 \right) \to \left( 2 \right) \Rightarrow {x^2} + {\left( {2x - 1} \right)^2} = 2\\
 \Leftrightarrow {x^2} + 4{x^2} - 4x + 1 = 2\\
 \Leftrightarrow 5{x^2} - 4x - 1 = 0\\
 \Leftrightarrow \left( {x - 1} \right)\left( {5x + 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x =  - \dfrac{1}{5}
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
y = 1\\
y =  - \dfrac{7}{5}
\end{array} \right.\\
 \Rightarrow \left( {x;y} \right) = \left( {1;1} \right),\left( { - \dfrac{1}{5}; - \dfrac{7}{5}} \right)
\end{array}$