Easy way to decompose $\frac{1}{X^{3}\cdot (X-2)^{3}}$ into partial fractions?

Question

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 ?

Answer 1

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 $X-1$.

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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: $$ B = \lim_{t\to 1}(1-t)^2\left(g(t)-\frac{1}{8(1-t)^3}\right) = \lim_{t\to 1}\frac{7+4t+t^2}{8(1+t)^3}=\frac{3}{16}.\tag{5}$$

Answer 2

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 $(a-b)^3 = a^3-b^3+3a^2b-3ab^2$ $$\frac{1}{x^3(x-2)^3} = -\frac{1}{8}\left(\frac {1}{x^3}-\frac{1}{(x-2)^3}+\frac{3}{x^2(x-2)}-\frac{3}{x(x-2)^2}\right)$$

Answer 3

You can also do this as follows. If $R(x)$ is a rational function that tends to zero at infinity, then expanding $R(x)$ around each of its singularities in the complex plane, keeping only the singular parts of the expansion and adding up all the terms from all these expansions will yield the partial fraction expansion.

The reason why this works is simple, if you consider the difference between $R(x)$ and the sum of all the singular parts of the expansions, then the resulting function only has removable singularities, as you've subtracted precisely the singular behavior at all the singular points. Therefore what we have is a polynomial. Since $R(x)$ and all the terms subtracted from $R(x)$ tend to zero at infinity, this polynomial is identical to zero.

In this case there is a singularity at $x = 0$ and at $x = 2$. Expanding around $x = 0$ amounts to expanding the factor $\frac{1}{(x-2)^3}$ in positive powers of $x$, for our purpose we only need to keep terms up to $x^2$. This is easily done:

$$\frac{1}{(x-2)^3} = -\frac{1}{8}\left(1-\frac{x}{2}\right)^{-3}= -\frac{1}{8}\left[1+\frac{3}{2}x+\frac{3}{2}x^2+\cdots\right]$$

So, the part of the partial fraction expansion coming from the singularity at $x = 0$, is given by:

$$S_1(x) = -\frac{1}{8 x^3} - \frac{3}{16 x^2} - \frac{3}{16 x}$$

Expanding around the singularity at $x = 2$ yields the remaining part of the partial fraction expansion. Putting $x = t + 2$, we see that we need to expand the function

$$\widetilde{R}(t) = \frac{1}{t^3(t+2)^3}$$

around $t = 0$. But this looks similar to what we've done above to obtain $S_1(x)$. Note that $\widetilde{R}(t) = R(-t)$, therefore the singular terms from the expansion around $x = 2$ are given by:

$$\widetilde{S}_2(t) = S_1(-t) = \frac{1}{8 t^3} - \frac{3}{16 t^2} + \frac{3}{16 t}$$

The partial fraction expansion is thus given by:

$$-\frac{1}{8 x^3} - \frac{3}{16 x^2} - \frac{3}{16 x} + \frac{1}{8 (x-2)^3} - \frac{3}{16 (x-2)^2} + \frac{3}{16 (x-2)}$$