Воспользуемся формулой сокращённого умножения: \({a^2} + 2ab + {b^2} = {\left( {a + b} \right)^2}.\)
\({x^2} + 8x + 16 = {x^2} + 2 \cdot x \cdot 4 + {4^2} = {\left( {x + 4} \right)^2}.\)
\(\left( {x-2} \right)\left( {{x^2} + 8x + 16} \right) = 7\left( {x + 4} \right)\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left( {x-2} \right){\left( {x + 4} \right)^2}-7\left( {x + 4} \right) = 0\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\left( {x + 4} \right)\left( {\left( {x-2} \right)\left( {x + 4} \right)-7} \right) = 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}x + 4 = 0,\\\left( {x-2} \right)\left( {x + 4} \right)-7 = 0\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\left[ \begin{array}{l}x = -4,\\{x^2} + 2x-15 = 0.\end{array} \right.\)
Решим второе уравнение последней совокупности:
\({x^2} + 2x-15 = 0;\,\,\,\,\,\,D = {2^2}-4 \cdot 1 \cdot \left( {-15} \right) = 64;\,\,\,\,\,\,\,\left[ \begin{array}{l}x = \dfrac{{-2-8}}{2} = -5,\\x = \dfrac{{-2 + 8}}{2} = 3.\end{array} \right.\)
Тогда: \(\left[ \begin{array}{l}x = -4,\\{x^2} + 2x-15 = 0\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}x = -4,\\x = -5,\\x = 3.\end{array} \right.\)
Ответ: –5; –4; 3.