Question Number 131613 by mohammad17 last updated on 06/Feb/21
$$\frac{{z}^{\mathrm{3}} +\mathrm{8}}{{z}^{\mathrm{4}} +\mathrm{4}{z}^{\mathrm{2}} +\mathrm{16}} \\ $$
Answered by Dwaipayan Shikari last updated on 06/Feb/21
$$\frac{{z}^{\mathrm{3}} +\mathrm{8}}{{z}^{\mathrm{4}} +\mathrm{4}{z}^{\mathrm{2}} +\mathrm{16}}=\frac{\left({z}+\mathrm{2}\right)\left({z}^{\mathrm{2}} −\mathrm{2}{z}+\mathrm{4}\right)}{\left({z}^{\mathrm{2}} −\mathrm{2}{z}+\mathrm{4}\right)\left({z}^{\mathrm{2}} +\mathrm{2}{z}+\mathrm{4}\right)}=\frac{{z}+\mathrm{2}}{{z}^{\mathrm{2}} +\mathrm{2}{z}+\mathrm{4}} \\ $$