Решите систему уравнений: x^3-y^3=98; x-y=2.

Ответ:

\[\left\{ \begin{matrix} x^{3} - y^{3} = 98 \\ x - y = 2\ \ \ \ \ \ \ \\ \end{matrix}\text{\ \ \ \ \ } \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} (x - y)\left( x^{2} + xy + y^{2} \right) = 98 \\ x - y = 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix}\text{\ \ \ } \right.\ \]

\[\left\{ \begin{matrix} 2 \cdot \left( x^{2} + xy + y^{2} \right) = 98\ \ \ \ \ \ |\ :2 \\ x = 2 + y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix}\text{\ \ } \right.\ \text{\ \ \ \ \ }\]

\[\left\{ \begin{matrix} x^{2} + xy + y^{2} = 49 \\ x = 2 + y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix}\text{\ \ \ \ } \right.\ \]

\[\left\{ \begin{matrix} (2 + y)^{2} + y(2 + y) + y^{2} = 49 \\ x = 2 + y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix}\text{\ \ \ \ } \right.\ \]

\[3y^{2} + 6y - 45 = 0\ \ \ \ \ \ \ \ \ \ \ \ |\ :3\]

\[y^{2} + 2y - 15 = 0\]

\[y_{1} + y_{2} = - 2,\ \ y_{1}y_{2} = - 15\]

\[\left\{ \begin{matrix} y = - 5 \\ x = - 3 \\ \end{matrix}\text{\ \ \ \ \ } \right.\ \]

\[\left\{ \begin{matrix} y = 3 \\ x = 5 \\ \end{matrix} \right.\ \]

\[Ответ:( - 3;\ - 5),\ (5;3).\]