partial fraction decomposition of $\frac{k^4}{(a\, k^3-1)^2}$

Question

I have to perform complex partial fraction decomposition of the following term:

$$\frac{k^4}{(a \, k^3-1)^2}$$

where $a$ is a real positive number.

and I would like to know if it is possible to reduce it to a sum of fractions of the type $\frac{A}{k\pm z}$,$\frac{B \, k}{k^2\pm z}$,$\frac{C \, k}{k^2-y}$ or similar. Where $z$ is a complex number and $y$ is a real number.

If it is not possible to reduce it to the kind of fraction I listed above other type of decomposition might work too.

Any hint on the process, or any reference would be nice.

Thanks in advance

Answer 1

I would start by making the change of variable $ak^{3} = x^{3}$, which transforms the expression into

$$ a^{3/4}\frac{x^{4}}{(x^{3} - 1)^{2}}.$$

Now we factorise the denominator. This is easy since we have a difference of two cubes.

$$ x^{3} - 1 = (x-1)(x^{2} + x + 1).$$

Notice that the quadratic is irreducible over the reals. Then (ignoring the constant $a^{3/4}$) we have

$$ \frac{x^{4}}{(x-1)^{2}(x^{2} + x + 1)} = \frac{A}{x - 1} + \frac{B}{(x-1)^{2}} + \frac{Cx + D}{x^{2} + x + 1} + \frac{Ex + F}{(x^{2} + x + 1)^{2}}.$$

Answer 2

Set $a=b^3$ and denominator will be factored as $$\left(b^3 k^3-1\right)^2=(b k-1)^2 \left(b^2 k^2+b k+1\right)^2$$ Then you decompose in partial fractions $$\frac{e+f k}{b^2 k^2+b k+1}+\frac{g+h k}{\left(b^2 k^2+b k+1\right)^2}+\frac{c}{b k-1}+\frac{d}{(b k-1)^2}$$ where $c,d,e,f,g,h$ are unknowns and must be obtained solving a six equations linear system. First you add all the fractions and consider the numerator of the result

$b^5 c k^5+b^4 c k^4+b^4 d k^4+b^4 e k^4+b^4 f k^5+b^3 c k^3+2 b^3 d k^3-b^3 e k^3-b^3 f k^4-b^2 c k^2+3 b^2 d k^2+b^2 g k^2+b^2 h k^3-b c k+2 b d k-b e k-b f k^2-2 b g k-2 b h k^2-c+d+e+f k+g+h k$

Numerator must be identical to $k^4$ so coefficients (from degree 0 up to 5 must be $\{0,0,0,0,1,0\}$

The system has quite nice solutions $$c= \frac{2}{9 b^4},d= \frac{1}{9 b^4},e= \frac{4}{9 b^4},f= -\frac{2}{9 b^3},g= -\frac{1}{3 b^4},h= 0$$ Therefore the starting fraction can be written as $$\frac{1}{9b^4}\left(-\frac{2 (b k-2)}{b^2 k^2+b k+1}-\frac{3}{\left(b^2 k^2+b k+1\right)^2}+\frac{2}{b k-1}+\frac{1}{(b k-1)^2}\right)$$

with $b={a}^\frac{1}{3}$

Answer 3

Decompose MR.wolframalpha's solution, on $\omega^3=1$ $\dfrac{1}{3(2\omega+1)}\bigg(\dfrac1{x-\omega^2}-\dfrac 1{x-\omega}\bigg)+\dfrac1{x-1}+\dfrac {1}{3(x-1)^2}+1$