$\begin{align} & \text{If }x\text{ is so small that we can neglect terms in }{{x}^{4}}\text{ and higher power of }x,\text{ } \\ & \text{show that }\sqrt[3]{\frac{1-2x}{1+2x}}=1-\frac{4}{5}x+\frac{8}{9}{{x}^{2}}-\frac{170}{81}{{x}^{3}} \\\end{align}$
$\begin{align} & \sqrt[3]{\frac{1-2x}{1+2x}}={{\left( \frac{1-2x}{1+2x} \right)}^{\tfrac{1}{3}}}={{(1-2x)}^{\tfrac{1}{3}}}{{(1+2x)}^{-\tfrac{1}{3}}} \\ & {{(1-2x)}^{\tfrac{1}{3}}}=1+\tfrac{1}{3}(-2x)+\frac{\tfrac{1}{3}(-\tfrac{2}{3})}{2}{{(-2x)}^{2}}+\frac{\tfrac{1}{3}(-\tfrac{2}{3})(-\tfrac{5}{3})}{6}{{(-2x)}^{3}}+-- \\ & {{(1-2x)}^{\tfrac{1}{3}}}=1-\frac{2x}{3}-\frac{4{{x}^{2}}}{9}-\frac{40{{x}^{3}}}{81}+--- \\ & {{(1+2x)}^{-\tfrac{1}{3}}}=1-\tfrac{1}{3}(2x)+\frac{-\tfrac{1}{3}(-\tfrac{4}{3})}{2}{{(2x)}^{2}}+\frac{-\tfrac{1}{3}(-\tfrac{4}{3})(-\tfrac{7}{3})}{6}{{(2x)}^{3}}+-- \\ & {{(1+2x)}^{-\tfrac{1}{3}}}=1-\frac{2x}{3}+\frac{8{{x}^{2}}}{9}-\frac{112{{x}^{3}}}{81}+--- \\ & {{(1-2x)}^{\tfrac{1}{3}}}{{(1+2x)}^{-\tfrac{1}{3}}}=\left( 1-\frac{2x}{3}-\frac{4{{x}^{2}}}{9}-\frac{40{{x}^{3}}}{81}+--- \right)\times \left( 1-\frac{2x}{3}+\frac{8{{x}^{2}}}{9}-\frac{112{{x}^{3}}}{81}+--- \right) \\ & {{(1-2x)}^{\tfrac{1}{3}}}{{(1+2x)}^{-\tfrac{1}{3}}}=1-\frac{2x}{3}+\frac{8{{x}^{2}}}{9}-\frac{2x}{3}+\frac{4{{x}^{2}}}{9}-\frac{16{{x}^{3}}}{27}-\frac{4{{x}^{2}}}{9}+\frac{8{{x}^{3}}}{27}-\frac{40{{x}^{3}}}{81} \\ & {{(1-2x)}^{\tfrac{1}{3}}}{{(1+2x)}^{-\tfrac{1}{3}}}=1-\frac{4x}{3}+\frac{8{{x}^{2}}}{9}-\frac{64{{x}^{3}}}{27}+---- \\\end{align}$