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 ?

We may consider that:
$$ \frac{1}{t-1}-\frac{1}{t+1} = \frac{2}{(t-1)(t+1)}\tag{1} $$
so:
$$\begin{eqnarray*} \frac{8}{(t-1)^3 (t+1)^3} &=& \frac{1}{(t-1)^3}-\frac{3}{(t-1)^2(t+1)}+\frac{3}{(t-1)(t+1)^2}-\frac{1}{(t+1)^3}\\&=&\frac{1}{(t-1)^3}-\frac{1}{(t+1)^3}+\frac{3}{2}\frac{2}{(t-1)(t+1)}\left(\frac{1}{t+1}-\frac{1}{t-1}\right)\\&=&\frac{1}{(t-1)^3}-\frac{1}{(t+1)^3}-\frac{3}{2}\left(\frac{1}{t-1}-\frac{1}{t+1}\right)^2\\&=&\frac{1}{(t-1)^3}-\frac{1}{(t+1)^3}-\frac{3/2}{(t-1)^2}-\frac{3/2}{(t+1)^2}+\frac{3}{(t-1)(t+1)}\\&=&\frac{1}{(t-1)^3}-\frac{1}{(t+1)^3}-\frac{3/2}{(t-1)^2}-\frac{3/2}{(t+1)^2}+\frac{3/2}{t-1}-\frac{3/2}{t+1}\tag{2}\end{eqnarray*}$$
and now it is enough to replace $t$ with $X-1$.

Alternative approach. Assuming that
$$\frac{1}{(1-t)^3(1+t)^3}=\frac{A}{(1-t)^3}+\frac{B}{(1-t)^2}+\frac{C}{1-t}+\frac{D}{1+t}+\frac{E}{(1+t)^2}+\frac{F}{(1+t)^3}\tag{3}$$
we have $A=F,B=E$ and $C=D$ since the LHS is an even function.
So, it is enough to find $A,B,C$. If we set $g(t)=\frac{1}{(1-t^2)^3}$, we have:
$$ A = \lim_{t\to 1}(1-t)^3 g(t) = \lim_{t\to 1}\frac{1}{(1+t)^3}=\frac{1}{8}\tag{4}$$
and $A+B+C+D+E+F = g(0)=1$, so it is enough to find $B$, for instance through:
$$ B = \lim_{t\to 1}(1-t)^2\left(g(t)-\frac{1}{8(1-t)^3}\right) = \lim_{t\to 1}\frac{7+4t+t^2}{8(1+t)^3}=\frac{3}{16}.\tag{5}$$
