<s>[INST] Easy way to decompose $\frac{1}{X^{3}\cdot (X-2)^{3}}$ into partial fractions? Is there any easy way or shortcut  to decompose $$\frac{1}{X^{3}\cdot (X-2)^{3}}$$ into partial fractions  ?   because dealing with the usual way of replacing and giving values to $X$ is too clumsy in this case , so I was wondering if there is any quick way to find the decomposition of this rational function ?
 [/INST] Rewriting it,
$$\left(\frac{1}{x(x-2)}\right)^3$$
Using Partial fractions,
$$\left(\frac{1}{x(x-2)}\right)^3 = \left(-\frac{1}{2}\left(\frac{1}{x}-\frac{1}{x-2}\right)\right)^3$$
Applying $(a-b)^3 = a^3-b^3+3a^2b-3ab^2$
$$\frac{1}{x^3(x-2)^3} = -\frac{1}{8}\left(\frac {1}{x^3}-\frac{1}{(x-2)^3}+\frac{3}{x^2(x-2)}-\frac{3}{x(x-2)^2}\right)$$
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