а) \(\frac{\sqrt[n]{3^5}}{(\sqrt[n]{3})^{-5}} ⋅ 3^\(\frac{5}{n}\) = \(\frac{3^{\frac{5}{n}}}{3^{-\frac{5}{n}}} ⋅ 3^{\frac{5}{n}}\) = 3^{\frac{5}{n} + \frac{5}{n} + \frac{5}{n}}\) = 3^{\frac{15}{n}}
б) (27a6)\(\frac{1}{3}\) = (33a6)\(\frac{1}{3}\) = 3a2
Ответ: а) 3^{\frac{15}{n}}; б) 3a2