Question Number 131613 by mohammad17 last updated on 06/Feb/21
![((z^3 +8)/(z^4 +4z^2 +16))](https://www.tinkutara.com/question/Q131613.png)
$$\frac{{z}^{\mathrm{3}} +\mathrm{8}}{{z}^{\mathrm{4}} +\mathrm{4}{z}^{\mathrm{2}} +\mathrm{16}} \\ $$
Answered by Dwaipayan Shikari last updated on 06/Feb/21
![((z^3 +8)/(z^4 +4z^2 +16))=(((z+2)(z^2 −2z+4))/((z^2 −2z+4)(z^2 +2z+4)))=((z+2)/(z^2 +2z+4))](https://www.tinkutara.com/question/Q131616.png)
$$\frac{{z}^{\mathrm{3}} +\mathrm{8}}{{z}^{\mathrm{4}} +\mathrm{4}{z}^{\mathrm{2}} +\mathrm{16}}=\frac{\left({z}+\mathrm{2}\right)\left({z}^{\mathrm{2}} −\mathrm{2}{z}+\mathrm{4}\right)}{\left({z}^{\mathrm{2}} −\mathrm{2}{z}+\mathrm{4}\right)\left({z}^{\mathrm{2}} +\mathrm{2}{z}+\mathrm{4}\right)}=\frac{{z}+\mathrm{2}}{{z}^{\mathrm{2}} +\mathrm{2}{z}+\mathrm{4}} \\ $$