Question Number 94521 by Abdulrahman last updated on 19/May/20
$$\mathrm{if}\:\mathrm{a}^{\mathrm{10}} +\mathrm{a}^{\mathrm{5}} +\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{then}\:\mathrm{a}^{\mathrm{2005}} +\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2005}} }=? \\ $$$$\mathrm{a}:\:\mathrm{a}^{\mathrm{10}} +\mathrm{a}^{\mathrm{11}} \:\:\:\:\mathrm{b}:\mathrm{a}^{\mathrm{10}} +\mathrm{a}^{\mathrm{5}} \:\:\:\mathrm{c}:\mathrm{3}\left(\mathrm{a}^{\mathrm{10}} +\mathrm{a}^{\mathrm{5}} \right)\:\:\mathrm{d}:\mathrm{0} \\ $$$$\mathrm{with}\:\mathrm{steps}? \\ $$
Commented by peter frank last updated on 19/May/20
$${check}\:\mathrm{92839} \\ $$
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+a^5 +a^(10) =0 ⇒1+a^5 +(a^5 )^2 =0 ⇒((1−(a^5 )^3 )/(1−a^5 )) =0 ⇒a^(15) =1 and a^5 ≠0 the roots of a^(15) =0 are z_k =e^(i((2kπ)/(15))) k∈[[0,14]] a^(2005) +a^(−2005) =(e^(i((2kπ(2005))/(15))) + e^(−i((2kπ(2005))/(15))) ) =2cos(((2kπ(2005))/(15))) =2 cos(((4010kπ)/(15))) =2cos(267π+ (π/3)) =2cos(π+(π/3)) =−2cos((π/3)) =−2×(1/2) =−1](Q94621.png)
$$\mathrm{1}+\mathrm{a}^{\mathrm{5}} \:+\mathrm{a}^{\mathrm{10}} \:=\mathrm{0}\:\Rightarrow\mathrm{1}+\mathrm{a}^{\mathrm{5}} \:+\left(\mathrm{a}^{\mathrm{5}} \right)^{\mathrm{2}} \:=\mathrm{0}\:\Rightarrow\frac{\mathrm{1}−\left(\mathrm{a}^{\mathrm{5}} \right)^{\mathrm{3}} }{\mathrm{1}−\mathrm{a}^{\mathrm{5}} }\:=\mathrm{0}\:\Rightarrow\mathrm{a}^{\mathrm{15}} \:=\mathrm{1}\:\mathrm{and}\:\mathrm{a}^{\mathrm{5}} \:\neq\mathrm{0} \\ $$$$\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{a}^{\mathrm{15}} \:=\mathrm{0}\:\mathrm{are}\:\mathrm{z}_{\mathrm{k}} =\mathrm{e}^{\mathrm{i}\frac{\mathrm{2k}\pi}{\mathrm{15}}} \:\:\:\:\mathrm{k}\in\left[\left[\mathrm{0},\mathrm{14}\right]\right] \\ $$$$\mathrm{a}^{\mathrm{2005}} \:+\mathrm{a}^{−\mathrm{2005}} \:=\left(\mathrm{e}^{\mathrm{i}\frac{\mathrm{2k}\pi\left(\mathrm{2005}\right)}{\mathrm{15}}} \:+\:\mathrm{e}^{−\mathrm{i}\frac{\mathrm{2k}\pi\left(\mathrm{2005}\right)}{\mathrm{15}}} \right) \\ $$$$=\mathrm{2cos}\left(\frac{\mathrm{2k}\pi\left(\mathrm{2005}\right)}{\mathrm{15}}\right)\:=\mathrm{2}\:\mathrm{cos}\left(\frac{\mathrm{4010k}\pi}{\mathrm{15}}\right)\:=\mathrm{2cos}\left(\mathrm{267}\pi+\:\frac{\pi}{\mathrm{3}}\right) \\ $$$$=\mathrm{2cos}\left(\pi+\frac{\pi}{\mathrm{3}}\right)\:=−\mathrm{2cos}\left(\frac{\pi}{\mathrm{3}}\right)\:=−\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}\:=−\mathrm{1} \\ $$$$ \\ $$