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Question Number 94521 by Abdulrahman last updated on 19/May/20

$$\mathrm{if}{\mathrm{a}}^{10}+{\mathrm{a}}^5+1=0$$$$\mathrm{then}{\mathrm{a}}^{2005}+\frac{1}{{\mathrm{a}}^{2005}}=?$$$$\mathrm{a}:{\mathrm{a}}^{10}+{\mathrm{a}}^{11}\mathrm{b}:{\mathrm{a}}^{10}+{\mathrm{a}}^5\mathrm{c}:3({\mathrm{a}}^{10}+{\mathrm{a}}^5)\mathrm{d}:0$$$$\mathrm{with}\mathrm{steps}?$$

Commented by peter frank last updated on 19/May/20

$$check92839$$

Commented by Abdulrahman last updated on 19/May/20

$$\mathrm{thanks}\mathrm{alot}$$

Commented by i jagooll last updated on 19/May/20

$$\mathrm{answer}\mathrm{B}$$

Answered by Abdulrahman last updated on 19/May/20

$$\mathrm{please}\mathrm{solve}$$

Answered by i jagooll last updated on 19/May/20

Commented by i jagooll last updated on 19/May/20

this is mr john's answer

Answered by mathmax by abdo last updated on 20/May/20

$$1+{\mathrm{a}}^5+{\mathrm{a}}^{10}=0\Rightarrow 1+{\mathrm{a}}^5+{\left({\mathrm{a}}^5\right)}^2=0\Rightarrow \frac{1-{\left({\mathrm{a}}^5\right)}^3}{1-{\mathrm{a}}^5}=0\Rightarrow {\mathrm{a}}^{15}=1\mathrm{and}{\mathrm{a}}^5\ne 0$$$$\mathrm{the}\mathrm{roots}\mathrm{of}{\mathrm{a}}^{15}=0\mathrm{are}{\mathrm{z}}_{\mathrm{k}}={\mathrm{e}}^{\mathrm{i}\frac{2\mathrm{k}\pi }{15}}\mathrm{k}\in \left[[0,14]\right]$$$${\mathrm{a}}^{2005}+{\mathrm{a}}^{-2005}=({\mathrm{e}}^{\mathrm{i}\frac{2\mathrm{k}\pi \left(2005\right)}{15}}+{\mathrm{e}}^{-\mathrm{i}\frac{2\mathrm{k}\pi \left(2005\right)}{15}})$$$$=2\cos \left(\frac{2\mathrm{k}\pi \left(2005\right)}{15}\right)=2\cos \left(\frac{4010\mathrm{k}\pi }{15}\right)=2\cos (267\pi +\frac{\pi }{3})$$$$=2\cos (\pi +\frac{\pi }{3})=-2\cos \left(\frac{\pi }{3}\right)=-2\times \frac{1}{2}=-1$$$$$$

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