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