Answer
$(2x-1)\sqrt[3]{2x}$
Work Step by Step
Simplify each radical (when needed) by factoring the radicand such that one of the factors is a perfect cube:
$\sqrt[3]{16x^4}-\sqrt[3]{2x}\\
=\sqrt[3]{8x^3\cdot 2x}-\sqrt[3]{2x}\\
=\sqrt[3]{2^3x^3\cdot 2x}-\sqrt[3]{2x}\\
=\sqrt[3]{(2x)^3\cdot 2x}-\sqrt[3]{2x}$
Use the rule $\sqrt[n]{a\cdot b}=\sqrt[n]{a} \cdot \sqrt[n]{b}$ then simplify:
$=\sqrt[3]{(2x)^3}\cdot \sqrt[3]{2x}-\sqrt[3]{2x}$
$=2x\cdot \sqrt[3]{2x}-\sqrt[3]{2x}$
$=2x\sqrt[3]{2x}-\sqrt[3]{2x}$
Factor out $\sqrt[3]{2x}$.
$=\sqrt[3]{2x}(2x-1)$
$=(2x-1)\sqrt[3]{2x}$
Hence, the correct answer is $(2x-1)\sqrt[3]{2x}$.
