Answer
$$x = \left\{ { - \frac{1}{{243}},\frac{1}{{3125}}} \right\}$$
Work Step by Step
$$\eqalign{
& {x^{ - 2/5}} - 2{x^{ - 1/5}} - 15 = 0 \cr
& {\text{Rewrite using the property }}{\left( {{a^m}} \right)^n} = {a^{mn}} \cr
& {\left( {{x^{ - 1/5}}} \right)^2} - 2\left( {{x^{ - 1/5}}} \right) - 15 = 0 \cr
& {\text{Factor the trinomial}} \cr
& \left( {{x^{ - 1/5}} - 5} \right)\left( {{x^{ - 1/5}} + 3} \right) = 0 \cr
& {\text{Use the zero - factor property}} \cr
& {x^{ - 1/5}} - 5 = 0{\text{ or }}{x^{ - 1/5}} + 3 = 0 \cr
& {\text{Solving both equations}} \cr
& {x^{ - 1/5}} = 5{\text{ or }}{x^{ - 1/5}} = - 3 \cr
& \frac{1}{{{x^{1/5}}}}{\text{ = 5 or }}\frac{1}{{{x^{1/5}}}} = - 3{\text{ }} \cr
& {x^{1/5}} = \frac{1}{5}{\text{ or }}{x^{1/5}} = - \frac{1}{3} \cr
& x = \frac{1}{{3125}}{\text{ or }}x = - \frac{1}{{243}} \cr} $$
