1. **State the problem:** Simplify the cube root expression $$\sqrt[3]{\frac{-324}{2}}$$. 2. **Rewrite the fraction inside the cube root:** $$\sqrt[3]{\frac{-324}{2}} = \sqrt[3]{-162}$$. 3. **Express -162 as a product of its prime factors:** $$-162 = -1 \times 2 \times 3^4$$. 4. **Use the property of cube roots:** $$\sqrt[3]{a \times b} = \sqrt[3]{a} \times \sqrt[3]{b}$$. 5. **Separate the cube root:** $$\sqrt[3]{-1 \times 2 \times 3^4} = \sqrt[3]{-1} \times \sqrt[3]{2} \times \sqrt[3]{3^4}$$. 6. **Simplify each cube root:** - $$\sqrt[3]{-1} = -1$$ because $(-1)^3 = -1$. - $$\sqrt[3]{3^4} = 3^{\frac{4}{3}} = 3^{1 + \frac{1}{3}} = 3 \times \sqrt[3]{3}$$. 7. **Combine all parts:** $$-1 \times \sqrt[3]{2} \times 3 \times \sqrt[3]{3} = -3 \times \sqrt[3]{2} \times \sqrt[3]{3} = -3 \times \sqrt[3]{2 \times 3} = -3 \times \sqrt[3]{6}$$. **Final answer:** $$\boxed{-3 \sqrt[3]{6}}$$