## Intermediate Algebra for College Students (7th Edition)

$\dfrac{5\sqrt[3]{2x}}{x}$
RECALL: For any real number $a$, $\sqrt[3]{a^3}= a$ Rationalize the denominator by multiplying $\sqrt[3]{2x}$ to both the numerator and the denominator. Simplify using the rule above to obtain: $\require{cancel} =\dfrac{10 \cdot \sqrt[3]{2x}}{\sqrt[3]{4x^2} \cdot \sqrt[3]{2x}} \\=\dfrac{10\sqrt[3]{2x}}{\sqrt[3]{8x^3}} \\=\dfrac{10\sqrt[3]{2x}}{\sqrt[3]{2^3x^3}} \\=\dfrac{10\sqrt[3]{2x}}{2x} \\=\dfrac{5\cancel{10}\sqrt[3]{2x}}{\cancel{2}x} \\=\dfrac{5\sqrt[3]{2x}}{x}$