1. Stating the problem: Simplify the expression $$\sqrt[3]{8} \times \sqrt[3]{16} \div \sqrt[3]{32}$$. 2. Use the property of cube roots: $$\sqrt[3]{a} \times \sqrt[3]{b} = \sqrt[3]{a \times b}$$ and $$\frac{\sqrt[3]{a}}{\sqrt[3]{b}} = \sqrt[3]{\frac{a}{b}}$$. 3. Combine the cube roots: $$\sqrt[3]{8} \times \sqrt[3]{16} \div \sqrt[3]{32} = \frac{\sqrt[3]{8 \times 16}}{\sqrt[3]{32}} = \sqrt[3]{\frac{8 \times 16}{32}}$$. 4. Simplify inside the cube root: $$\frac{8 \times 16}{32} = \frac{128}{32}$$. 5. Simplify the fraction: $$\frac{128}{32} = \cancel{\frac{128}{32}}^{4}$$ (since 128 divided by 32 equals 4). 6. So the expression becomes: $$\sqrt[3]{4}$$. 7. Final answer: $$\sqrt[3]{4}$$. This is the simplest form since 4 is not a perfect cube.