Đáp án:
$\left[ \begin{array}{l}
x = y = 0\\
x = y = \pm \sqrt 2
\end{array} \right.$
Giải thích các bước giải:
$\begin{array}{l}
\left\{ \begin{array}{l}
{x^2} - xy + {y^2} = 2\\
{y^3} = x + y
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
\left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right) = 2\left( {x + y} \right)\\
{y^3} = x + y
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{x^3} + {y^3} = 2{y^3}\\
{y^3} = x + y
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
{x^3} = {y^3}\\
{y^3} - y = x
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x = y\\
{y^3} - y - x = 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x = y\\
{y^3} - 2y = 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x = y\\
y\left( {{y^2} - 2} \right) = 0
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = y = 0\\
x = y = \pm \sqrt 2
\end{array} \right.
\end{array}$
