5．若函数$f(x) = \frac{\text{e}^{x} + a}{x^{3}}$在$(0, + \infty)$上存在极值,则$a$的取值范围为\\\\\\.
解:由题知$f(x) = \frac{\text{e}^{x} + a}{x^{3}}$在$(0, + \infty)$上存在极值,
即$f'(x)$在$(0, + \infty)$上存在变号零点,
所以$f'(x) = \frac{(x - 3)\text{e}^{x} - 3a}{x^{4}}$,
设函数$g(x) = (x - 3)\text{e}^{x} - 3a$,
即$g(x)$在$(0, + \infty)$上存在变号零点,
则$g'(x) = (x - 2)\text{e}^{x}$,
当$0 < x < 2$时,$g'(x) < 0$,$g(x)$单调递减;
当$x > 2$时,$g'(x) > 0$,$g(x)$单调递增.
因为$x \rightarrow + \infty$时,$g(x) \rightarrow + \infty$,
故只需$g(x)_{\text{min}} = g(2) = - \text{e}^{2} - 3a < 0$即可,
即$a > - \frac{\text{e}^{2}}{3}$.
故答案为:$\left( - \frac{\text{e}^{2}}{3}, + \infty \right)$.