Рациональные уравнения. Задача 17

ОТВЕТ: \(-3;\;\; \pm \sqrt 2 .\)

\({x^3} + 3x{}^2-2x-6 = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{x^2}\left( {x + 3} \right)-2\left( {x + 3} \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \)

\( \Leftrightarrow \,\,\,\,\,\,\,\left( {x + 3} \right)\left( {{x^2}-2} \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x + 3 = 0,}\\{{x^2}-2 = 0}\end{array}\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,} \right.\,\left[ {\begin{array}{*{20}{c}}{x = -3,\,\,\,\,}\\{x = -\sqrt 2 }\\{x = \sqrt 2 .\,}\end{array}} \right.\)

Ответ:  \(-3;\;\; \pm \sqrt 2 .\)