Воспользуемся формулой сокращённого умножения: \({a^2} + 2ab + {b^2} = {\left( {a + b} \right)^2}.\)
\({x^2} + 6x + 9 = {x^2} + 2 \cdot x \cdot 3 + {3^2} = {\left( {x + 3} \right)^2}.\)
\(x\left( {{x^2} + 6x + 9} \right) = 4\left( {x + 3} \right)\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,x{\left( {x + 3} \right)^2}-4\left( {x + 3} \right) = 0\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\left( {x + 3} \right)\left( {x\left( {x + 3} \right)-4} \right) = 0\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}x + 3 = 0,\\x\left( {x + 3} \right)-4 = 0\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}x = -3,\\{x^2} + 3x-4 = 0.\end{array} \right.\)
Решим второе уравнение последней совокупности:
\({x^2} + 3x-4 = 0;\,\,\,\,\,\,D = {3^2}-4 \cdot 1 \cdot \left( {-4} \right) = 25;\,\,\,\,\,\,\,\left[ \begin{array}{l}x = \dfrac{{-3-5}}{2} = -4,\\x = \dfrac{{-3 + 5}}{2} = 1.\end{array} \right.\)
Тогда: \(\left[ \begin{array}{l}x = -3,\\{x^2} + 3x-4 = 0\end{array} \right.\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\left[ \begin{array}{l}x = -3,\\x = -4,\\x = 1.\end{array} \right.\)
Ответ: –4; –3; 1.