Answer
$\dfrac{\sqrt[3]{3}}{10}+\sqrt[3]{\dfrac{24}{125}}=\dfrac{\sqrt[3]{3}}{2}$
Work Step by Step
$\dfrac{\sqrt[3]{3}}{10}+\sqrt[3]{\dfrac{24}{125}}$
Rewrite the second term as $\dfrac{\sqrt[3]{8\cdot3}}{\sqrt[3]{125}}$ and simplify it:
$\dfrac{\sqrt[3]{3}}{10}+\sqrt[3]{\dfrac{24}{125}}=\dfrac{\sqrt[3]{3}}{10}+\dfrac{\sqrt[3]{8\cdot3}}{\sqrt[3]{125}}=\dfrac{\sqrt[3]{3}}{10}+\dfrac{2\sqrt[3]{3}}{5}=...$
Evaluate the sum and simplify if possible:
$...=\Big(\dfrac{1+2\cdot2}{10}\Big)\sqrt[3]{3}=\dfrac{5}{10}\sqrt[3]{3}=\dfrac{\sqrt[3]{3}}{2}$
