\(\left( {x-5} \right){\left( {x + 4} \right)^2} \ge 0.\)
Решим неравенство методом интервалов:
\(\left( {x-5} \right){\left( {x + 4} \right)^2} = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x-5 = 0,\,\,\,\,\,}\\{{{\left( {x + 4} \right)}^2} = 0}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = 5,\,\,\,}\\{x = -4.}\end{array}} \right.\)
Следовательно, решение исходного неравенства: \(x\, \in \,\left\{ {-4} \right\} \cup \left[ {5; + \infty } \right).\)
Ответ: \(\{ -4\} \cup \left[ {5;\; + \infty } \right).\)