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Question Number 108480 by I want to learn more last updated on 17/Aug/20

Answered by Rasheed.Sindhi last updated on 17/Aug/20

      x^2 +x+1=0  ,  x^(500) +(1/x^(500) )=?  x^2 +x+1=0⇒x(x+(1/x)+1)=0  x≠0 ⇒ x+(1/x)=−1   ( x+(1/x))^2 =(−1)^2       x^2 +(1/x^2 )=1−2=−1 ..............(i)  (x+(1/x))^3 =(−1)^3    x^3 +(1/x^3 )+3(x+(1/x))=−1    x^3 +(1/x^3 )+3(−1)=−1   x^3 +(1/x^3 )=2...................(ii)  (i)×(ii)  x^5 +(1/x^5 )+x+(1/x)=−2  x^5 +(1/x^5 )+(−1)=−2  x^5 +(1/x^5 )=−1...................(iii)  (x^5 +(1/x^5 ))^2 =(−1)^2   x^(10) +(1/x^(10) )=−1  x^n +(1/x^n )=−1  ⇒(x^n +(1/x^n ))^2 =(−1)^2   ⇒x^(2n) +(1/x^(2n) )=1−2=−1       ⟨x^n +(1/x^n )=−1⇒x^(2n) +(1/x^(2n) )=−1⟩_(−)      ^(−)   Similarly        ⟨  x^n +(1/x^n )=2⇒x^(3n) +(1/x^(3n) )=2  ⟩_(−)      ^(−)     ....  ......

x2+x+1=0,x500+1x500=?x2+x+1=0x(x+1x+1)=0x0x+1x=1(x+1x)2=(1)2x2+1x2=12=1..............(i)(x+1x)3=(1)3x3+1x3+3(x+1x)=1x3+1x3+3(1)=1x3+1x3=2...................(ii)(i)×(ii)x5+1x5+x+1x=2x5+1x5+(1)=2x5+1x5=1...................(iii)(x5+1x5)2=(1)2x10+1x10=1xn+1xn=1(xn+1xn)2=(1)2x2n+1x2n=12=1xn+1xn=1x2n+1x2n=1Similarlyxn+1xn=2x3n+1x3n=2..........

Commented by I want to learn more last updated on 17/Aug/20

Thanks sir,  i appreciate

Thankssir,iappreciate

Commented by udaythool last updated on 17/Aug/20

supper

supper

Answered by 1549442205PVT last updated on 17/Aug/20

From the hypothesis we get  x^2 +x+1=0 Δ=1−4=−3=3i^2   ⇒x=((−1±i(√3))/2)=cos((2π)/3)±isin((2π)/3)  ⇒x^(500) =(cos((2π)/3)±isin((2π)/3))^(500)   =cos((1000π)/3)±sin((1000π)/3)=cos(333π+(π/3))±isin(333π+(π/3))  =−cos(π/3)∓isin(π/3)=−(1/2)∓i((√3)/2)  =((−1∓i(√3))/2)(1)  ⇒(1/x^(500) )=(2/(−1∓i(√3)))=((2(−1±i(√3)))/((−1)^2 −(i(√3))^2 ))=((2(−1±i(√3)))/(1−3i^2 ))  =((2(−1±i(√3)))/4)=((−1±i(√3))/2)(2)  Adding up two the eqialities (1) (2)  we get x^(500) +(1/x^(500) )=((−1∓i(√3)+−1±i(√3))/2)=−1  Thus,x^(500) +(1/x^(500) )=((−2)/2)=−1.

Fromthehypothesiswegetx2+x+1=0Δ=14=3=3i2x=1±i32=cos2π3±isin2π3x500=(cos2π3±isin2π3)500=cos1000π3±sin1000π3=cos(333π+π3)±isin(333π+π3)=cosπ3isinπ3=12i32=1i32(1)1x500=21i3=2(1±i3)(1)2(i3)2=2(1±i3)13i2=2(1±i3)4=1±i32(2)Addinguptwotheeqialities(1)(2)wegetx500+1x500=1i3+1±i32=1Thus,x500+1x500=22=1.

Commented by I want to learn more last updated on 17/Aug/20

Thanks sir

Thankssir

Answered by Dwaipayan Shikari last updated on 17/Aug/20

x^2 +x+1=0  x=e^((2iπ)/3)  or e^(−((2iπ)/3))   x^(500) =e^((1000iπ)/3) =e^(−((2πi)/3)) =−((1/2)+((√3)/2)i)  (1/x^(500) )=e^((−1000πi)/3) =e^((2iπ)/3) =−(1/2)+((√3)/2)i  x^(500) +(1/x^(500) )=−1  Or  x^(500) =e^(−((1000iπ)/3))   (1/x^(500) )=e^((1000iπ)/3)   x^(500) +(1/x^(500) )=−1

x2+x+1=0x=e2iπ3ore2iπ3x500=e1000iπ3=e2πi3=(12+32i)1x500=e1000πi3=e2iπ3=12+32ix500+1x500=1Orx500=e1000iπ31x500=e1000iπ3x500+1x500=1

Commented by udaythool last updated on 17/Aug/20

excellent

excellent

Commented by I want to learn more last updated on 17/Aug/20

Thanks sir

Thankssir

Answered by mnjuly1970 last updated on 17/Aug/20

sol :x^2 +x+1=0⇒x^3 =1  x^(500) +(1/x^(500) )=(x^3 )^(166)  .x^2 +((1/x^3 ))^(166) .(1/x^2 )  =x^2 +(1/x^2 )=(−1)^2 −2=−1♣♣♣

sol:x2+x+1=0x3=1x500+1x500=(x3)166.x2+(1x3)166.1x2=x2+1x2=(1)22=1

Commented by I want to learn more last updated on 17/Aug/20

Thanks sir

Thankssir

Answered by Rasheed.Sindhi last updated on 17/Aug/20

x^2 +x+1=0  x=((−1±(√(1−4)))/2)=((−1±i(√3))/2)  x=ω,ω^2   x^(500) +(1/x^(500) )=ω^(500) +(1/ω^(500) )  (ω^3 )^(166) .ω^2 +(1/((ω^3 )^(166) .ω^2 ))  =ω^2 +(1/ω^2 )×(ω/ω)=ω^2 +(ω/ω^3 )=ω^2 +ω=−1

x2+x+1=0x=1±142=1±i32x=ω,ω2x500+1x500=ω500+1ω500(ω3)166.ω2+1(ω3)166.ω2=ω2+1ω2×ωω=ω2+ωω3=ω2+ω=1

Commented by I want to learn more last updated on 17/Aug/20

Thanks sir

Thankssir

Answered by floor(10²Eta[1]) last updated on 17/Aug/20

x^2 =−(x+1)  x^4 =(−(x+1))^2 =x^2 +2x+1=−x−1+2x+1=x  x^5 =x^4 .x=x.x=x^2 =−x−1  x^(10) =(−(x+1))^2 =x^2 +2x+1=−x−1+2x+1=x  x^(10) =x∴x^(500) =x^(50) =(x^(10) )^5 =x^5 =−x−1=x^2   ⇒x^(500) +(1/x^(500) )=x^2 +(1/x^2 )=((x^4 +1)/x^2 )=((x+1)/(−(x+1)))=−1

x2=(x+1)x4=((x+1))2=x2+2x+1=x1+2x+1=xx5=x4.x=x.x=x2=x1x10=((x+1))2=x2+2x+1=x1+2x+1=xx10=xx500=x50=(x10)5=x5=x1=x2x500+1x500=x2+1x2=x4+1x2=x+1(x+1)=1

Commented by Rasheed.Sindhi last updated on 17/Aug/20

e^x cellent approach!

excellentapproach!

Commented by I want to learn more last updated on 17/Aug/20

Thanks sir

Thankssir

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