\(\left( {{x^2}-10x + 25} \right)\left( {x + 6} \right) \ge 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,{\left( {x-5} \right)^2}\left( {x + 6} \right) \ge 0.\)
Решим неравенство методом интервалов:
\({\left( {x-5} \right)^2}\left( {x + 6} \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{{{\left( {x-5} \right)}^2} = 0,}\\{x + 6 = 0\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{x = 5,\,\,\,}\\{x = -6.}\end{array}} \right.\)
Следовательно, решение исходного неравенства: \(x\, \in \,\left[ {-6; + \infty } \right).\)
Ответ: \(\left[ {-6;\; + \infty } \right).\)