Question Number 208814 by Ismoiljon_008 last updated on 23/Jun/24
Commented by Ismoiljon_008 last updated on 23/Jun/24
$${Help}\:{me}\:{please} \\ $$
Commented by Ismoiljon_008 last updated on 23/Jun/24
$${If}\:{x}^{\mathrm{4}} +{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}=\mathrm{0},\:{then}\:{x}^{\mathrm{71}} −{x}^{\mathrm{61}} +{x}^{\mathrm{51}} +{x}^{\mathrm{41}} +\mathrm{1}=? \\ $$
Answered by Rasheed.Sindhi last updated on 23/Jun/24
$${If}\:{x}^{\mathrm{4}} +{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}=\mathrm{0},\:{then}\:{x}^{\mathrm{71}} −{x}^{\mathrm{61}} +{x}^{\mathrm{51}} +{x}^{\mathrm{41}} +\mathrm{1}=? \\ $$$$\:{x}^{\mathrm{4}} +{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}=\mathrm{0} \\ $$$$\:\Rightarrow\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{4}} +{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$${x}^{\mathrm{5}} −\mathrm{1}=\mathrm{0}\Rightarrow{x}^{\mathrm{5}} =\mathrm{1} \\ $$$$\:{x}^{\mathrm{71}} −{x}^{\mathrm{61}} +{x}^{\mathrm{51}} +{x}^{\mathrm{41}} +\mathrm{1} \\ $$$$=\left({x}^{\mathrm{5}} \right)^{\mathrm{14}} \centerdot{x}−\left({x}^{\mathrm{5}} \right)^{\mathrm{12}} \centerdot{x}+\left({x}^{\mathrm{5}} \right)^{\mathrm{10}} \centerdot{x}+\left({x}^{\mathrm{5}} \right)^{\mathrm{8}} \centerdot{x}+\mathrm{1} \\ $$$$=\left(\mathrm{1}\right)^{\mathrm{14}} \centerdot{x}−\left(\mathrm{1}\right)^{\mathrm{12}} \centerdot{x}+\left(\mathrm{1}\right)^{\mathrm{10}} \centerdot{x}+\left(\mathrm{1}\right)^{\mathrm{8}} \centerdot{x}+\mathrm{1} \\ $$$$={x}−{x}+{x}+{x}+\mathrm{1}=\mathrm{2}{x}+\mathrm{1} \\ $$
Commented by Ismoiljon_008 last updated on 23/Jun/24
$${Thank}\:{you}\:{very}\:{much} \\ $$