Answer
$\sqrt[3]{\dfrac{16}{27}}-\dfrac{\sqrt[3]{54}}{6}=\dfrac{\sqrt[3]{2}}{6}$
Work Step by Step
$\sqrt[3]{\dfrac{16}{27}}-\dfrac{\sqrt[3]{54}}{6}$
Rewrite the first term as $\dfrac{\sqrt[3]{8\cdot2}}{\sqrt[3]{27}}$ and the second term as $\dfrac{\sqrt[3]{27\cdot2}}{6}$:
$\sqrt[3]{\dfrac{16}{27}}-\dfrac{\sqrt[3]{54}}{6}=\dfrac{\sqrt[3]{8\cdot2}}{\sqrt[3]{27}}-\dfrac{\sqrt[3]{27\cdot2}}{6}=...$
Simplify both terms:
$...=\dfrac{2\sqrt[3]{2}}{3}-\dfrac{3\sqrt[3]{2}}{6}=\dfrac{2\sqrt[3]{2}}{3}-\dfrac{\sqrt[3]{2}}{2}=...$
Evaluate the substraction and simplify if possible:
$...\Big(\dfrac{2}{3}-\dfrac{1}{2}\Big)\sqrt[3]{2}=\Big(\dfrac{4-3}{6}\Big)\sqrt[3]{2}=\dfrac{\sqrt[3]{2}}{6}$
