1. **State the problem:** Simplify the expression $$\sqrt[3]{\frac{18}{128}}$$. 2. **Recall the cube root property:** The cube root of a fraction is the fraction of the cube roots, i.e., $$\sqrt[3]{\frac{a}{b}} = \frac{\sqrt[3]{a}}{\sqrt[3]{b}}$$. 3. **Apply the property:** $$\sqrt[3]{\frac{18}{128}} = \frac{\sqrt[3]{18}}{\sqrt[3]{128}}$$ 4. **Factor numerator and denominator:** - 18 factors as $$2 \times 3^2$$ - 128 factors as $$2^7$$ 5. **Rewrite cube roots:** $$\frac{\sqrt[3]{2 \times 3^2}}{\sqrt[3]{2^7}} = \frac{\sqrt[3]{2} \times \sqrt[3]{3^2}}{\sqrt[3]{2^7}} = \frac{\sqrt[3]{2} \times \sqrt[3]{3^2}}{\sqrt[3]{2^7}}$$ 6. **Simplify the fraction inside the cube root for 2:** $$\frac{\sqrt[3]{2}}{\sqrt[3]{2^7}} = \sqrt[3]{\frac{2}{2^7}} = \sqrt[3]{2^{1-7}} = \sqrt[3]{2^{-6}}$$ 7. **Rewrite the expression:** $$\sqrt[3]{2^{-6}} \times \sqrt[3]{3^2} = \sqrt[3]{2^{-6} \times 3^2}$$ 8. **Simplify inside the cube root:** $$\sqrt[3]{\frac{3^2}{2^6}} = \sqrt[3]{\frac{9}{64}}$$ 9. **Since 64 is $$4^3$$, and 9 is $$3^2$$, rewrite as:** $$\frac{\sqrt[3]{9}}{\sqrt[3]{64}} = \frac{\sqrt[3]{9}}{4}$$ 10. **Final simplified form:** $$\frac{\sqrt[3]{9}}{4}$$ **Answer:** $$\boxed{\frac{\sqrt[3]{9}}{4}}$$