Answer
$\color{blue}{\dfrac{-25\sqrt[3]{9}}{18}}$
Work Step by Step
Simplify the denominators to obtain:
$=\dfrac{-4}{\sqrt[3]{3}} + \dfrac{1}{\sqrt[3]{8(3)}} - \dfrac{2}{\sqrt[3]{27(3)}}
\\=\dfrac{-4}{\sqrt[3]{3}} + \dfrac{1}{2\sqrt[3]{3}}- \dfrac{2}{3\sqrt[3]{3}}$
Make the expressions similar by using their LCD of $6\sqrt[3]{3}$ to obtain:
$=\dfrac{-4(6)}{\sqrt[3]{3}(6)}+ \dfrac{1(3)}{2\sqrt[3]{3}(3)}-\dfrac{2(2)}{3\sqrt[3]{3}(2)}
\\=\dfrac{-24}{6\sqrt[3]{3}}+\dfrac{3}{6\sqrt[3]{3}}-\dfrac{4}{6\sqrt[3]{3}}$
Perform the operations to the numerators and copy the denominator to obtain:
$=\dfrac{-24+3-4}{6\sqrt[3]{3}}
\\=\dfrac{-25}{6\sqrt[3]{3}}$
Rationalize the denominator by multiplying $\sqrt[3]{9}$ to both the numerator and the denominator to obtain:
$=\dfrac{-25(\sqrt[3]{9})}{6\sqrt[3]{3}(\sqrt[3]{9})}
\\=\dfrac{-25\sqrt[3]{9}}{6(3)}
\\=\color{blue}{\dfrac{-25\sqrt[3]{9}}{18}}$
