1. The problem is to verify if $$\sqrt[6]{\frac{9}{16}a^2b^4} = \sqrt[3]{\frac{3}{4}ab^2}$$ is correct. 2. Recall the rule for radicals: $$\sqrt[n]{x^m} = x^{\frac{m}{n}}$$. 3. Rewrite each side using exponents: Left side: $$\sqrt[6]{\frac{9}{16}a^2b^4} = \left(\frac{9}{16}a^2b^4\right)^{\frac{1}{6}} = \left(\frac{9}{16}\right)^{\frac{1}{6}} a^{\frac{2}{6}} b^{\frac{4}{6}} = \left(\frac{9}{16}\right)^{\frac{1}{6}} a^{\frac{1}{3}} b^{\frac{2}{3}}$$ Right side: $$\sqrt[3]{\frac{3}{4}ab^2} = \left(\frac{3}{4}ab^2\right)^{\frac{1}{3}} = \left(\frac{3}{4}\right)^{\frac{1}{3}} a^{\frac{1}{3}} b^{\frac{2}{3}}$$ 4. Compare the coefficients: Left coefficient: $$\left(\frac{9}{16}\right)^{\frac{1}{6}} = \left(\frac{3^2}{4^2}\right)^{\frac{1}{6}} = \left(\frac{3}{4}\right)^{\frac{2}{6}} = \left(\frac{3}{4}\right)^{\frac{1}{3}}$$ Right coefficient: $$\left(\frac{3}{4}\right)^{\frac{1}{3}}$$ 5. Since the coefficients and the variable parts are equal, the original equation is correct. Final answer: Yes, $$\sqrt[6]{\frac{9}{16}a^2b^4} = \sqrt[3]{\frac{3}{4}ab^2}$$ is correct.