1. **State the problem:** Simplify the expressions $$\sqrt[4]{25x^2}$$ and $$\sqrt[6]{81g^3}$$. 2. **Recall the formula:** The nth root of a product is the product of the nth roots: $$\sqrt[n]{a^m} = a^{\frac{m}{n}}$$. 3. **Simplify the first expression:** $$\sqrt[4]{25x^2} = \sqrt[4]{25} \cdot \sqrt[4]{x^2} = 25^{\frac{1}{4}} \cdot x^{\frac{2}{4}} = 25^{\frac{1}{4}} \cdot x^{\frac{1}{2}}$$ Since $$25 = 5^2$$, then $$25^{\frac{1}{4}} = (5^2)^{\frac{1}{4}} = 5^{\frac{2}{4}} = 5^{\frac{1}{2}} = \sqrt{5}$$ So the first expression simplifies to $$\sqrt{5} \cdot x^{\frac{1}{2}} = \sqrt{5} \cdot \sqrt{x} = \sqrt{5x}$$. 4. **Simplify the second expression:** $$\sqrt[6]{81g^3} = \sqrt[6]{81} \cdot \sqrt[6]{g^3} = 81^{\frac{1}{6}} \cdot g^{\frac{3}{6}} = 81^{\frac{1}{6}} \cdot g^{\frac{1}{2}}$$ Since $$81 = 3^4$$, then $$81^{\frac{1}{6}} = (3^4)^{\frac{1}{6}} = 3^{\frac{4}{6}} = 3^{\frac{2}{3}}$$ We can write $$3^{\frac{2}{3}} = (3^{\frac{1}{3}})^2 = (\sqrt[3]{3})^2$$. So the second expression simplifies to $$(\sqrt[3]{3})^2 \cdot \sqrt{g}$$. **Final answers:** $$\sqrt[4]{25x^2} = \sqrt{5x}$$ $$\sqrt[6]{81g^3} = (\sqrt[3]{3})^2 \cdot \sqrt{g}$$