Answer
$2x\sqrt[3]{2}-\sqrt[3]{x^2}$
Work Step by Step
RECALL:
(1) Distributive Property:
For any real numbers a, b, and c,
$a(b-c)=ab-ac$
(2) For any real numbers real numbers a and b within the domain,
$\sqrt[n]{a} \cdot \sqrt[n]{b}=\sqrt[n]{ab}$
Use rule (1) above to obtain:
$=\sqrt[3]{x}(\sqrt[3]{16x^2}) - \sqrt[3]{x} \cdot \sqrt[3]{x}$
Use rule (2) above to obtain:
$=\sqrt[3]{16x^3} - \sqrt[3]{x^2}$
Factor the first radicand so that at least one factor is a perfect cube to obtain:
$=\sqrt[3]{8x^3(2)} - \sqrt[3]{x^2}
\\=\sqrt[3]{(2x)^3(2)} - \sqrt[3]{x^2}$
Simplify to obtain:
$=2x\sqrt[3]{2}-\sqrt[3]{x^2}$
