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Question Number 92839 by i jagooll last updated on 09/May/20

$$\mathrm{let}\mathrm{a}\mathrm{is}\mathrm{complex}\mathrm{number}\mathrm{such}$$$$\mathrm{that}{\mathrm{a}}^{10}+{\mathrm{a}}^5+1=0.$$$$\mathrm{find}{\mathrm{a}}^{2005}+\frac{1}{{\mathrm{a}}^{2005}}?$$

Answered by john santu last updated on 09/May/20

$$\mathrm{set}{a}^5=w\Rightarrow w^2+w+1=0$$$$w^2+w+1=\frac{w^3-1}{w-1}=0$$$$\Rightarrow a^{10}+a^5+1=\frac{a^{15}-1}{a^5-1}=0$$$$a^{15}=1\Rightarrow \mathrm{since}2005=133\times 15+10$$$$a^{2005}={\left(a^{15}\right)}^{133}\times a^{10}=1\times a^{10}=a^{10}$$$$\mathrm{then}{a}^{2005}+\frac{1}{a^{2005}}=a^{10}+\frac{1}{a^{10}}$$$$=a^{10}+\frac{a^{15}}{a^{10}}=a^{10}+a^5$$$$=-1[a^{10}+a^5+1=0]$$

Commented by i jagooll last updated on 09/May/20

$$\mathrm{thank}\mathrm{you}$$

Commented by peter frank last updated on 09/May/20

$$thankyou$$

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