multiply: $4xsqrt3{4x^{2}}(2sqrt3{32x^{2}} - xsqrt3{2x})$ \n$32x^{2}sqrt3{2x}-8sqrt3{x^{2}}$…

Explanation:

Step1: Distribute the term

$4x\sqrt[3]{4x^{2}}\times2\sqrt[3]{32x^{2}}-4x\sqrt[3]{4x^{2}}\times x\sqrt[3]{2x}$

Step2: Multiply the coefficients and radicals separately for the first - part

For $4x\sqrt[3]{4x^{2}}\times2\sqrt[3]{32x^{2}}$, the coefficient is $4\times2 = 8$, and $\sqrt[3]{4x^{2}}\times\sqrt[3]{32x^{2}}=\sqrt[3]{4x^{2}\times32x^{2}}=\sqrt[3]{128x^{4}}=\sqrt[3]{64\times2x^{3}\times x}=4x\sqrt[3]{2x}$. So $4x\sqrt[3]{4x^{2}}\times2\sqrt[3]{32x^{2}}=8x\times4x\sqrt[3]{2x}=32x^{2}\sqrt[3]{2x}$.

Step3: Multiply the coefficients and radicals separately for the second - part

For $4x\sqrt[3]{4x^{2}}\times x\sqrt[3]{2x}$, the coefficient is $4\times1 = 4$, and $\sqrt[3]{4x^{2}}\times\sqrt[3]{2x}=\sqrt[3]{4x^{2}\times2x}=\sqrt[3]{8x^{3}} = 2x$. So $4x\sqrt[3]{4x^{2}}\times x\sqrt[3]{2x}=4x\times x\times2x=8x^{3}$.

Answer:

$32x^{2}\sqrt[3]{2x}-8x^{3}$