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

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