Answer
$2\sqrt[3] {4} + 1$
Work Step by Step
To get rid of radicals in the denominator, multiply both numerator and denominator by a factor that will make the radical in the denominator an integer. In this exercise, that factor would be $\sqrt[3] {2^2}$:
$\dfrac{4 + \sqrt[3] {2}}{\sqrt[3] {2}} \cdot \dfrac{\sqrt[3] {2^2}}{\sqrt[3] {2^2}}$
Use distributive property:
$=\dfrac{4(\sqrt[3] {2^2}) + \sqrt[3] {2}(\sqrt[3] {2^2})}{\sqrt[3] {2}(\sqrt[3] {2^2})}$
Multiply to simplify:
$=\dfrac{4\sqrt[3] {4} + \sqrt[3] {2^3}}{\sqrt[3] {2^3}}$
Simplify radicals:
$=\dfrac{4\sqrt[3] {4} + 2}{2}$
Simplify the fraction by dividing both numerator and denominator by their greatest common factor, $2$:
$2\sqrt[3] {4} + 1$
