I have to perform complex partial fraction decomposition of the following term: (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 (A)/(k+-z), (B k)/(k^2+-z), (Ck)/(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

I would start by making the change of variable $a k^{3 }= x^{3 }$, which transforms the expression into
$a^{3 / 4 }\frac{x^{4 }}{( x^{3 }- 1 {)}^{2 }}.$
The denominator is now factored. Given that there is only a two-cube difference, this is simple.
$x^{3 }- 1 = ( x - 1 ) ( x^{2 }+ x + 1 ) .$
The quadratic is irreducible over the reals, as you can see. Then (ignoring the constant $a^{3 / 4 }$) we have
$\frac{x^{4 }}{( x - 1 {)}^{2 }( x^{2 }+ x + 1 ) }= \frac{A}{x - 1 }+ \frac{B}{( x - 1 {)}^{2 }}+ \frac{C x + D }{x^{2 }+ x + 1 }+ \frac{E x + F }{( x^{2 }+ x + 1 {)}^{2 }}.$

Set $a = b^3$and denominator will be factored as
${( b^3{k}^3- 1 ) }^2= ( b k - 1 {)}^2{( b^2{k}^2+ b k + 1 ) }^2$
After that, you split into partial fractions.
$\frac{e + f k }{b^2{k}^2+ b k + 1 }+ \frac{g + h k }{{( b^2{k}^2+ b k + 1 ) }^2}+ \frac{c}{b k - 1 }+ \frac{d}{( b k - 1 {)}^2}$
where $c , d , e , f , g , h$ are unknowns and must be obtained solving a six equations linear system. Prior to considering the numerator of the result, you first sum all the fractions.
$b^{5}c k^5+ b^{4}c k^4+ b^{4}d k^4+ b^{4}e k^4+ b^{4}f k^5+ b^{3}c k^3+ 2 b^{3}d k^3- b^{3}e k^3- b^{3}f k^4- b^{2}c k^2+ 3 b^{2}d k^2+ b^{2}g k^2+ b^{2}h k^3- b c k + 2 b d k - b e k - b f k^2- 2 b g k - 2 b h k^2- c + d + e + f k + g + h k$
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 offers several lovely solutions.
$c = \frac{2}{9 b^4}, d = \frac{1}{9 b^4}, e = \frac{4}{9 b^4}, f = - \frac{2}{9 b^3}, g = - \frac{1}{3 b^4}, h = 0$
Hence, the initial fraction can be expressed as
$\frac{1}{9 b^4}( - \frac{2 ( b k - 2 ) }{b^2{k}^2+ b k + 1 }- \frac{3}{{( b^2{k}^2+ b k + 1 ) }^2}+ \frac{2}{b k - 1 }+ \frac{1}{( b k - 1 {)}^2})$
with $b = {a }^{\frac{1}{3}}$