Воспользуемся формулой сокращённого умножения: \({a^2} + 2ab + {b^2} = {\left( {a + b} \right)^2}.\)
\({x^2} + 2x + 1 = {x^2} + 2 \cdot x \cdot 1 + {1^2} = {\left( {x + 1} \right)^2}.\)
\(\left( {x-2} \right)\left( {{x^2} + 2x + 1} \right) = 4\left( {x + 1} \right)\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left( {x-2} \right){\left( {x + 1} \right)^2}-4\left( {x + 1} \right) = 0\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\left( {x + 1} \right)\left( {\left( {x-2} \right)\left( {x + 1} \right)-4} \right) = 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}x + 1 = 0,\\\left( {x-2} \right)\left( {x + 1} \right)-4 = 0\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}x = -1,\\{x^2}-x-6 = 0.\end{array} \right.\)
Решим второе уравнение последней совокупности:
\({x^2}-x-6 = 0;\,\,\,\,\,\,D = {\left( {-1} \right)^2}-4 \cdot 1 \cdot \left( {-6} \right) = 25;\,\,\,\,\,\,\,\left[ \begin{array}{l}x = \dfrac{{1-5}}{2} = -2,\\x = \dfrac{{1 + 5}}{2} = 3.\end{array} \right.\)
Тогда: \(\left[ \begin{array}{l}x = -1,\\{x^2}-x-6 = 0\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}x = -1,\\x = -2,\\x = 3.\end{array} \right.\)
Ответ: –2; –1; 3.