Воспользуемся формулой сокращённого умножения: \({a^2} + 2ab + {b^2} = {\left( {a + b} \right)^2}.\)
\({x^2} + 2x + 1 = {x^2} + 2 \cdot x \cdot 1 + {1^2} = {\left( {x + 1} \right)^2}.\)
\(x\left( {{x^2} + 2x + 1} \right) = 2\left( {x + 1} \right)\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x{\left( {x + 1} \right)^2}-2\left( {x + 1} \right) = 0\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\left( {x + 1} \right)\left( {x\left( {x + 1} \right)-2} \right) = 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}x + 1 = 0,\\x\left( {x + 1} \right)-2 = 0\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}x = -1,\\{x^2} + x-2 = 0.\end{array} \right.\)
Решим второе уравнение последней совокупности:
\({x^2} + x-2 = 0;\,\,\,\,\,\,D = {1^2}-4 \cdot 1 \cdot \left( {-2} \right) = 9;\,\,\,\,\,\,\,\left[ \begin{array}{l}x = \dfrac{{-1-3}}{2} = -2,\\x = \dfrac{{-1 + 3}}{2} = 1.\end{array} \right.\)
Тогда: \(\left[ \begin{array}{l}x = -1,\\{x^2} + x-2 = 0\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}x = -1,\\x = -2,\\x = 1.\end{array} \right.\)
Ответ: –2; –1; 1.