\({\left( {x-3} \right)^2}\left( {x + 1} \right) \le 0.\)
Решим неравенство методом интервалов:
\({\left( {x-3} \right)^2}\left( {x + 1} \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{{{\left( {x-3} \right)}^2} = 0,}\\{x + 1 = 0\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = 3,\,\,\,\,}\\{x = -1.}\end{array}} \right.\)
Следовательно, решение исходного неравенства: \(x\, \in \,\left( {-\infty ;-1} \right] \cup \left\{ 3 \right\}.\)
Ответ: \(\left( {-\infty ;\;-1} \right] \cup \{ 3\} .\)