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Since $e$ is small, the solution $x$ is close to $-1$. So write $x=-1+u$ and write your equation as $u(1-u)^{-1/3}=-e^{1/3}$. Then use the [Lagrange inversion formula][1]. [1]:http://en.wikipediaLagrange inversion formula.org/wiki/Formal_power_series#The_Lagrange_inversion_formula

Since $e$ is small, the solution $x$ is close to $-1$. So write $x=-1+u$ and write your equation as $u(1-u)^{-1/3}=-e^{1/3}$. Then use the [Lagrange inversion formula][1]. [1]:http://en.wikipedia.org/wiki/Formal_power_series#The_Lagrange_inversion_formula

Since $e$ is small, the solution $x$ is close to $-1$. So write $x=-1+u$ and write your equation as $u(1-u)^{-1/3}=-e^{1/3}$. Then use the Lagrange inversion formula.

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Pietro Majer
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Since $e$ is small, the solution $x$ is close to $-1$. So write $x=-1+u$ and write your equation as $u(1-u)^{-1/3}=-e^{1/3}$. Then use the [Lagrange inversion formula][1]. [1]:http://en.wikipedia.org/wiki/Formal_power_series#The_Lagrange_inversion_formula