1. **Simplify** the expression $$\sqrt[3]{\frac{125 \times x^{-4}}{x^2}} \times \sqrt{\frac{x^4}{25}}$$. 2. Apply laws of exponents inside the cube root: $$\sqrt[3]{\frac{125 \times x^{-4}}{x^2}} = \sqrt[3]{125 \times x^{-4-2}} = \sqrt[3]{125 \times x^{-6}}$$. 3. Inside the cube root, simplify the constants and powers: $$\sqrt[3]{125}=5, \text{ so } \sqrt[3]{125 \times x^{-6}} = 5 \times \sqrt[3]{x^{-6}}$$. 4. Simplify $$\sqrt[3]{x^{-6}}$$ by rewriting as $$x^{-6/3} = x^{-2}$$. 5. So the first part simplifies to $$5x^{-2}$$. 6. Now simplify the square root part: $$\sqrt{\frac{x^4}{25}} = \frac{\sqrt{x^4}}{\sqrt{25}} = \frac{x^{4/2}}{5} = \frac{x^2}{5}$$. 7. Multiply the two simplified parts: $$5x^{-2} \times \frac{x^2}{5} = \frac{5}{5} \times x^{-2} \times x^2 = 1 \times x^{(-2+2)} = x^{0} = 1$$. --- 8. **Solve** for $$x$$ in $$\log_6 1 + \frac{\log_6 216}{\log_6 36} = x$$. 9. Recall $$\log_6 1 = 0$$ because $$6^0=1$$. 10. Calculate $$\log_6 216$$. Since $$216 = 6^3$$, $$\log_6 216 = 3$$. 11. Calculate $$\log_6 36$$. Since $$36 = 6^2$$, $$\log_6 36 = 2$$. 12. Substitute values: $$x = 0 + \frac{3}{2} = \frac{3}{2} = 1.5$$. **Final answers:** 1. Simplified expression equals $$1$$. 2. Value of $$x$$ is $$\frac{3}{2}$$ or 1.5.