\(x\left( {{x^2} + 2x + 1} \right) = 2\left( {x + 1} \right)\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,x{\left( {x + 1} \right)^2}-2\left( {x + 1} \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,\left( {x + 1} \right)\left( {x\left( {x + 1} \right)-2} \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left( {x + 1} \right)\left( {{x^2} + x-2} \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x + 1 = 0,\,\,\,\,\,\,\,\,\,\,\,}\\{{x^2} + x-2 = 0}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = -1,}\\{x = -2,}\\{x = 1.\,\,\,}\end{array}} \right.\)
Ответ: –2; –1; 1.