I am trying to find the residues of the expression: $$f(z) = \frac{e^{z*t}(z^2-2)}{z^3-a\,z^2-2a}$$ but I don't reallly know how to do it, so far I've tried to use partial fractions and also find the roots of the denominator:
$\left\{\left\{z\to \frac{1}{3} \left(\sqrt[3]{a^3+3 \sqrt{3} \sqrt{2 a ^4+27 a^2}+27 a }+\frac{a^2}{\sqrt[3]{a ^3+3 \sqrt{3} \sqrt{2 a^4+27 a^2}+27 a }}+a \right)\right\},\left\{z\to -\frac{1}{6} \left(1-i \sqrt{3}\right) \sqrt[3]{a ^3+3 \sqrt{3} \sqrt{2 a ^4+27 a ^2}+27a }-\frac{\left(1+i \sqrt{3}\right) a^2}{6 \sqrt[3]{a^3+3 \sqrt{3} \sqrt{2 a^4+27 a^2}+27 a }}+\frac{a}{3}\right\},\left\{z\to -\frac{1}{6} \left(1+i \sqrt{3}\right) \sqrt[3]{\gamma ^3+3 \sqrt{3} \sqrt{2 a^4+27 a^2}+27 a}-\frac{\left(1-i \sqrt{3}\right) a^2}{6 \sqrt[3]{a ^3+3 \sqrt{3} \sqrt{2 a^4+27 a^2}+27 a}}+\frac{a}{3}\right\}\right\}$
, but I think I don't really get a usefull result, I am really knew in this kind of things so I don' t rellay know how to preceed, does anyone has a clue?, Thanks in advance