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МатематикаГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 345
Задание 345
\[\boxed{\mathbf{345}.}\]
\[x_{1};x_{2};x_{3} - корни\ \]
\[уравнения\ x^{3} + px + q.\]
\[\left\{ \begin{matrix} x_{1} + x_{2} + x_{3} = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \\ x_{1}x_{2} + x_{2}x_{3} + x_{1}x_{3} = p \\ x_{1}x_{2}x_{3} = - q\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\]
\[\ \left\{ \begin{matrix} x_{2} + x_{3} = - x_{1} \\ x_{1} + x_{3} = - x_{2} \\ x_{1} + x_{2} = - x_{3} \\ \end{matrix} \right.\ \]
\[\left( x_{1} + x_{2} + x_{3} \right)^{3} =\]
\[= x_{1}^{3} + x_{2}^{3} + x_{3}^{3} + 3x_{1}^{2}\left( x_{2} + x_{3} \right) +\]
\[+ 3x_{2}^{2}\left( x_{1} + x_{3} \right) +\]
\[+ 3x_{3}^{2}\left( x_{1} + x_{2} \right) + 6x_{1}x_{2}x_{3}\]
\[x_{1}^{3} + x_{2}^{3} + x_{3}^{3} =\]
\[= \left( \underset{= 0}{\overset{x_{1} + x_{2} + x_{3}}{︸}} \right)^{2} -\]
\[- 3x_{1}^{2} \cdot \left( - x_{1} \right) - 3x_{2}^{2}\left( - x_{2} \right) -\]
\[- 3x_{3}^{2} \cdot \left( - x_{3} \right) - 6x_{1}x_{2}x_{3}\]
\[\left( x_{1}^{3} + x_{2}^{3} + x_{3}^{3} \right) =\]
\[= 3 \cdot \left( x_{1}^{3} + x_{2}^{3} + x_{3}^{3} \right) - 6x_{1}x_{2}x_{3}\]
\[- 2 \cdot \left( x_{1}^{3} + x_{2}^{3} + x_{3}^{3} \right) = - 6x_{1}x_{2}x_{3}\text{\ \ \ \ }\]
\[x_{1}^{3} + x_{2}^{3} + x_{3}^{3} = 3x_{1}x_{2}x_{3}.\]
\[Что\ и\ требовалось\ доказать.\]
