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