Решите систему уравнений: x^3+y^3=7; x^2-xy+y^2=7.

Ответ:

\[\left\{ \begin{matrix} x^{3} + y^{3} = 7\ \ \ \ \ \ \ \ \ \ \\ x^{2} - xy + y^{2} = 7 \\ \end{matrix}\text{\ \ \ \ \ } \right.\ \]

\[\left\{ \begin{matrix} (x + y)\left( x^{2} - xy + y^{2} \right) = 7 \\ x^{2} - xy + y = 7\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix}\text{\ \ \ \ } \right.\ \]

\[\left\{ \begin{matrix} 7 \cdot (x + y) = 7\ \ \ \ \\ x^{2} - xy + y^{2} = 7 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x + y = 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - xy + y^{2} = 7 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x = 1 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \ (1 - y)^{2} - (1 - y)y + y^{2} = 7 \\ \end{matrix}\text{\ \ \ \ \ } \right.\ \]

\[\left\{ \begin{matrix} x = 1 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 1 - 2y + y^{2} - y + y^{2} + y^{2} = 7 \\ \end{matrix}\text{\ \ } \right.\ \]

\[\left\{ \begin{matrix} x = 1 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 3y^{2} - 3y - 6 = 0\ \ \ \ \ \ |\ :3 \\ \end{matrix} \right.\ \]

\[y^{2} - y - 2 = 0\]

\[y_{1} + y_{2} = 1;\ \ \ \ \ \ \ y_{1} = - 1\]

\[y_{1}y_{2} = - 2;\ \ \ \ \ \ \ \ \ y_{2} = 2\]

\[\left\{ \begin{matrix} y = - 1 \\ x - 2\ \ \ \ \\ \end{matrix}\text{\ \ \ \ \ \ } \right.\ \text{\ \ \ \ \ \ \ }\left\{ \begin{matrix} y = 2\ \ \ \\ x = - 1 \\ \end{matrix} \right.\ \]

\[Ответ:(2;\ - 1);\ \ ( - 1;2).\]