If $b$ is very small as compared to the value of $a$, so that the cube and other higher powers of $\frac{b}{a}$ can be neglected in the identity $\frac{1}{a-b}=\frac{1}{a-2 b}+\frac{1}{a-3 b}+\ldots . .+\frac{1}{a-n b}=\alpha n+\beta n^2+\gamma n^3$, then the value of $\gamma$ is :
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