Решение:
- Преобразуем выражение:
\(\sqrt[3]{81} = \sqrt[3]{3^4} = 3^{\frac{4}{3}}\)
\(\sqrt[6]{\frac{16}{6}} = \sqrt[6]{\frac{2^4}{2 \cdot 3}} = \sqrt[6]{\frac{2^3}{3}} = \sqrt[6]{\frac{2^3}{3^1}} = \left(\frac{2^3}{3^1}\right)^{\frac{1}{6}} = \frac{2^{\frac{3}{6}}}{3^{\frac{1}{6}}} = \frac{2^{\frac{1}{2}}}{3^{\frac{1}{6}}}\) - Теперь перемножим:
\(3^{\frac{4}{3}} \cdot \frac{2^{\frac{1}{2}}}{3^{\frac{1}{6}}} = 3^{\frac{4}{3} - \frac{1}{6}} \cdot 2^{\frac{1}{2}} = 3^{\frac{8}{6} - \frac{1}{6}} \cdot 2^{\frac{1}{2}} = 3^{\frac{7}{6}} \cdot 2^{\frac{1}{2}} = 3^{1 + \frac{1}{6}} \cdot 2^{\frac{1}{2}} = 3 \cdot 3^{\frac{1}{6}} \cdot 2^{\frac{3}{6}} = 3 \cdot \left(3 \cdot 2^3\right)^{\frac{1}{6}} = 3 \cdot \left(3 \cdot 8\right)^{\frac{1}{6}} = 3 \cdot \sqrt[6]{24}\)
Ответ: \(3 \sqrt[6]{24}\).