\(\left( {{x^2} + 6x + 9} \right){\left( {x-3} \right)^2} \le 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\left( {x + 3} \right)^2}{\left( {x-3} \right)^2} \le 0.\)
Решим неравенство методом интервалов:
\({\left( {x + 3} \right)^2}{\left( {x-3} \right)^2} = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{{{\left( {x + 3} \right)}^2} = 0,\,\,\,\,\,}\\{{{\left( {x-3} \right)}^2} = 0\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = -3,\,\,\,}\\{x = 3.\,\,\,\,\,\,}\end{array}} \right.\)
Следовательно, решение исходного неравенства: \(x\, \in \,\left\{ {-3;3} \right\}.\)
Ответ: \(\{ -3;\;3\} .\)