1. Stating the problem: Simplify the expression $$\sqrt[6]{\frac{9}{16}a^2b^4}$$. 2. Formula and rules: The sixth root of a product is the product of the sixth roots: $$\sqrt[6]{xy} = \sqrt[6]{x} \cdot \sqrt[6]{y}$$. 3. Apply the root to each factor: $$\sqrt[6]{\frac{9}{16}} \cdot \sqrt[6]{a^2} \cdot \sqrt[6]{b^4}$$ 4. Simplify the numeric root: $$\sqrt[6]{\frac{9}{16}} = \frac{\sqrt[6]{9}}{\sqrt[6]{16}} = \frac{9^{\frac{1}{6}}}{16^{\frac{1}{6}}}$$ 5. Express powers as fractional exponents: $$9^{\frac{1}{6}} = (3^2)^{\frac{1}{6}} = 3^{\frac{2}{6}} = 3^{\frac{1}{3}}$$ $$16^{\frac{1}{6}} = (2^4)^{\frac{1}{6}} = 2^{\frac{4}{6}} = 2^{\frac{2}{3}}$$ 6. So numeric part is: $$\frac{3^{\frac{1}{3}}}{2^{\frac{2}{3}}}$$ 7. Simplify variable parts: $$\sqrt[6]{a^2} = a^{\frac{2}{6}} = a^{\frac{1}{3}}$$ $$\sqrt[6]{b^4} = b^{\frac{4}{6}} = b^{\frac{2}{3}}$$ 8. Combine all parts: $$\frac{3^{\frac{1}{3}}}{2^{\frac{2}{3}}} \cdot a^{\frac{1}{3}} \cdot b^{\frac{2}{3}} = \frac{3^{\frac{1}{3}} a^{\frac{1}{3}} b^{\frac{2}{3}}}{2^{\frac{2}{3}}}$$ Final answer: $$\boxed{\frac{3^{\frac{1}{3}} a^{\frac{1}{3}} b^{\frac{2}{3}}}{2^{\frac{2}{3}}}}$$