Question Number 144334 by mathdanisur last updated on 24/Jun/21
$${Reduct}\:{it}:\:\:\frac{\boldsymbol{{z}}^{\mathrm{8}} \:+\:\boldsymbol{{z}}\:+\:\mathrm{1}}{\boldsymbol{{z}}^{\mathrm{5}} \:+\:\boldsymbol{{z}}\:+\:\mathrm{1}} \\ $$
Answered by Olaf_Thorendsen last updated on 24/Jun/21
$$\mathrm{Z}\:=\:\frac{{z}^{\mathrm{8}} +{z}+\mathrm{1}}{{z}^{\mathrm{5}} +{z}+\mathrm{1}} \\ $$$$\mathrm{Z}\:=\:{z}^{\mathrm{3}} −\frac{{z}^{\mathrm{4}} +{z}^{\mathrm{3}} −{z}−\mathrm{1}}{{z}^{\mathrm{5}} +{z}+\mathrm{1}} \\ $$$$\mathrm{Z}\:=\:{z}^{\mathrm{3}} −\frac{\left({z}+\mathrm{1}\right)\left({z}^{\mathrm{3}} −\mathrm{1}\right)}{{z}^{\mathrm{5}} +{z}+\mathrm{1}} \\ $$$$\mathrm{Z}\:=\:{z}^{\mathrm{3}} −\frac{\left({z}+\mathrm{1}\right)\left({z}−\mathrm{1}\right)\left({z}^{\mathrm{2}} +{z}+\mathrm{1}\right)}{{z}^{\mathrm{5}} +{z}+\mathrm{1}} \\ $$
Commented by mathdanisur last updated on 25/Jun/21
$${thanks}\:{Sir}\:{cool} \\ $$