1. **State the problem:** Multiply $$4 \times \sqrt[3]{4x^2} \times \left(2\sqrt[3]{32x^2} - x\sqrt[3]{2x}\right)$$ using the distributive property. 2. **Recall the distributive property:** $$a(b - c) = ab - ac$$. We will multiply $$4 \sqrt[3]{4x^2}$$ by each term inside the parentheses. 3. **Simplify each cube root:** - $$\sqrt[3]{4x^2} = \sqrt[3]{4} \times \sqrt[3]{x^2}$$ - $$\sqrt[3]{32x^2} = \sqrt[3]{32} \times \sqrt[3]{x^2} = 2 \times \sqrt[3]{4} \times \sqrt[3]{x^2}$$ (since $$32 = 2^5$$ and $$\sqrt[3]{32} = 2 \sqrt[3]{4}$$) - $$\sqrt[3]{2x} = \sqrt[3]{2} \times \sqrt[3]{x}$$ 4. **Multiply the first term:** $$4 \sqrt[3]{4x^2} \times 2 \sqrt[3]{32x^2} = 4 \times 2 \times \sqrt[3]{4x^2} \times \sqrt[3]{32x^2} = 8 \times \sqrt[3]{(4x^2)(32x^2)} = 8 \times \sqrt[3]{128x^4}$$ 5. **Simplify inside the cube root:** $$128 = 2^7$$, so $$\sqrt[3]{128x^4} = \sqrt[3]{2^7 x^4} = 2^{7/3} x^{4/3} = 2^{2 + 1/3} x^{4/3} = 4 \times 2^{1/3} x^{4/3}$$ 6. **So the first term becomes:** $$8 \times 4 \times 2^{1/3} x^{4/3} = 32 \times 2^{1/3} x^{4/3} = 32 x^{4/3} \sqrt[3]{2}$$ 7. **Multiply the second term:** $$4 \sqrt[3]{4x^2} \times (-x \sqrt[3]{2x}) = -4x \times \sqrt[3]{4x^2} \times \sqrt[3]{2x} = -4x \times \sqrt[3]{(4x^2)(2x)} = -4x \times \sqrt[3]{8x^3}$$ 8. **Simplify inside the cube root:** $$\sqrt[3]{8x^3} = 2x$$ 9. **So the second term becomes:** $$-4x \times 2x = -8x^2$$ 10. **Combine both terms:** $$32 x^{4/3} \sqrt[3]{2} - 8 x^2$$ 11. **Rewrite powers:** Note that $$x^{4/3} = x^{1 + 1/3} = x \times x^{1/3}$$, so the first term can be written as $$32 x^{1 + 1/3} \sqrt[3]{2} = 32 x^{5/3} \sqrt[3]{2}$$ **Final answer:** $$32 x^{5/3} \sqrt[3]{2x} - 8 x^3$$ This matches the second option in the multiple choice list.