Question Number 159860 by Khalmohmmad last updated on 21/Nov/21
Commented by MJS_new last updated on 22/Nov/21
$${t}+{t}^{−\mathrm{1}} =−\mathrm{1} \\ $$$${t}^{\mathrm{2}} +{t}+\mathrm{1}=\mathrm{0}\:\Rightarrow\:\mathrm{no}\:\mathrm{real}\:\mathrm{solution} \\ $$$${t}=−\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i} \\ $$$${t}^{\mathrm{1402}} ={t} \\ $$$${t}^{−\mathrm{1363}} ={t}^{−\mathrm{1}} =−\frac{\mathrm{1}}{\mathrm{2}}\mp\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i} \\ $$$$\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:−\mathrm{1} \\ $$
Answered by mindispower last updated on 22/Nov/21
$${u}+\frac{\mathrm{1}}{{u}}=−\mathrm{1} \\ $$$${u}^{\mathrm{2}} +{u}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{u}^{\mathrm{3}} =\mathrm{1} \\ $$$$\mathrm{1402}=\mathrm{3}.\mathrm{467}+\mathrm{1} \\ $$$$\mathrm{1363}=\mathrm{3}.\mathrm{454}+\mathrm{1} \\ $$$$\Rightarrow{u}^{\mathrm{1402}} +\frac{\mathrm{1}}{{u}^{\mathrm{1363}} }={u}+\frac{\mathrm{1}}{{u}}=−\mathrm{1} \\ $$