# 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

partial fraction decomposition of $\frac{{k}^{4}}{\left(a\phantom{\rule{thinmathspace}{0ex}}{k}^{3}-1{\right)}^{2}}$
I have to perform complex partial fraction decomposition of the following term:
$\frac{{k}^{4}}{\left(a\phantom{\rule{thinmathspace}{0ex}}{k}^{3}-1{\right)}^{2}}$
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 $\frac{A}{k±z}$,$\frac{B\phantom{\rule{thinmathspace}{0ex}}k}{{k}^{2}±z}$,$\frac{C\phantom{\rule{thinmathspace}{0ex}}k}{{k}^{2}-y}$ 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.
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yelashwag8
I would start by making the change of variable $a{k}^{3}={x}^{3}$, which transforms the expression into
${a}^{3/4}\frac{{x}^{4}}{\left({x}^{3}-1{\right)}^{2}}.$
Now we factorise the denominator. This is easy since we have a difference of two cubes.
${x}^{3}-1=\left(x-1\right)\left({x}^{2}+x+1\right).$
Notice that the quadratic is irreducible over the reals. Then (ignoring the constant ${a}^{3/4}$) we have
$\frac{{x}^{4}}{\left(x-1{\right)}^{2}\left({x}^{2}+x+1\right)}=\frac{A}{x-1}+\frac{B}{\left(x-1{\right)}^{2}}+\frac{Cx+D}{{x}^{2}+x+1}+\frac{Ex+F}{\left({x}^{2}+x+1{\right)}^{2}}.$
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Glenn Hopkins
Set $a={b}^{3}$ and denominator will be factored as
${\left({b}^{3}{k}^{3}-1\right)}^{2}=\left(bk-1{\right)}^{2}{\left({b}^{2}{k}^{2}+bk+1\right)}^{2}$
Then you decompose in partial fractions
$\frac{e+fk}{{b}^{2}{k}^{2}+bk+1}+\frac{g+hk}{{\left({b}^{2}{k}^{2}+bk+1\right)}^{2}}+\frac{c}{bk-1}+\frac{d}{\left(bk-1{\right)}^{2}}$
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
${b}^{5}c{k}^{5}+{b}^{4}c{k}^{4}+{b}^{4}d{k}^{4}+{b}^{4}e{k}^{4}+{b}^{4}f{k}^{5}+{b}^{3}c{k}^{3}+2{b}^{3}d{k}^{3}-{b}^{3}e{k}^{3}-{b}^{3}f{k}^{4}-{b}^{2}c{k}^{2}+3{b}^{2}d{k}^{2}+{b}^{2}g{k}^{2}+{b}^{2}h{k}^{3}-bck+2bdk-bek-bf{k}^{2}-2bgk-2bh{k}^{2}-c+d+e+fk+g+hk$
Numerator must be identical to ${k}^{4}$ so coefficients (from degree 0 up to 5 must be $\left\{0,0,0,0,1,0\right\}$
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
$\frac{1}{9{b}^{4}}\left(-\frac{2\left(bk-2\right)}{{b}^{2}{k}^{2}+bk+1}-\frac{3}{{\left({b}^{2}{k}^{2}+bk+1\right)}^{2}}+\frac{2}{bk-1}+\frac{1}{\left(bk-1{\right)}^{2}}\right)$
with $b={a}^{\frac{1}{3}}$