\({\left( {x-2} \right)^3}{\left( {x + 4} \right)^2} > 0.\)
Решим неравенство методом интервалов:
\({\left( {x-2} \right)^3}{\left( {x + 4} \right)^2} = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{{{\left( {x-2} \right)}^3} = 0,}\\{{{\left( {x + 4} \right)}^2} = 0}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = 2,\,\,\,}\\{x = -4.}\end{array}} \right.\)
Следовательно, решение исходного неравенства: \(x\, \in \,\left( {2; + \infty } \right).\)
Ответ: \(\left( {2;\; + \infty } \right).\)