1. **State the problem:** Simplify the expression $$\frac{\sqrt{32b^{10} c^6}}{\sqrt{9b^{12} c^4}}$$ assuming all variables represent positive numbers. 2. **Recall the property of radicals:** $$\frac{\sqrt{A}}{\sqrt{B}} = \sqrt{\frac{A}{B}}$$ for positive $A$ and $B$. 3. **Apply the property:** $$\frac{\sqrt{32b^{10} c^6}}{\sqrt{9b^{12} c^4}} = \sqrt{\frac{32b^{10} c^6}{9b^{12} c^4}}$$ 4. **Simplify inside the radical:** $$\frac{32b^{10} c^6}{9b^{12} c^4} = \frac{32}{9} \cdot b^{10-12} \cdot c^{6-4} = \frac{32}{9} b^{-2} c^{2}$$ 5. **Rewrite with positive exponents:** $$\frac{32}{9} \cdot \frac{c^{2}}{b^{2}}$$ 6. **Express the radical:** $$\sqrt{\frac{32}{9} \cdot \frac{c^{2}}{b^{2}}} = \sqrt{\frac{32 c^{2}}{9 b^{2}}}$$ 7. **Separate the radical:** $$\frac{\sqrt{32} \cdot \sqrt{c^{2}}}{\sqrt{9} \cdot \sqrt{b^{2}}}$$ 8. **Simplify each radical:** - $$\sqrt{32} = \sqrt{16 \cdot 2} = 4 \sqrt{2}$$ - $$\sqrt{c^{2}} = c$$ (since $c$ is positive) - $$\sqrt{9} = 3$$ - $$\sqrt{b^{2}} = b$$ (since $b$ is positive) 9. **Combine all parts:** $$\frac{4 \sqrt{2} \cdot c}{3 \cdot b} = \frac{4 c \sqrt{2}}{3 b}$$ **Final answer:** $$\boxed{\frac{4 c \sqrt{2}}{3 b}}$$