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Question Number 46110 by Tawa1 last updated on 21/Oct/18

Commented by math khazana by abdo last updated on 22/Oct/18

Δ=1−4=−3=(i(√3))^2 ⇒α=((1+i(√3))/2) and β=((1−i(√3))/2) ⇒α=e^(i(π/3)) and β =e^(−i(π/3)) ⇒α^(101) =e^(i((101)/3)π) = e^(i(((99 +2)/3))π) =e^(33iπ) e^(i((2π)/3)) =−j β^(107) = e^(−i((107)/3)π) =e^(−i((101)/3)π) e^(−i((6π)/3)) = e^(−i33π) e^(−i((2π)/3)) =−j^− ⇒α^(101) +β^(107) =−j−j^− =−(j+j^− ) =−(1+j+j^− ) +1=0+1 =1

Δ=1−4=−3=(i3)2⇒α=1+i32andβ=1−i32⇒α=eiπ3andβ=e−iπ3⇒α101=ei1013π=ei(99+23)π=e33iπei2π3=−jβ107=e−i1073π=e−i1013πe−i6π3=e−i33πe−i2π3=−j−⇒α101+β107=−j−j−=−(j+j−)=−(1+j+j−)+1=0+1=1

Commented by Tawa1 last updated on 22/Oct/18

God bless you sir

Godblessyousir

Answered by ajfour last updated on 21/Oct/18

x^3 +1 = (x+1)(x^2 −x+1) roots of x^3 +1=0 are −1, −ω^2 , −ω So (−ω)^(101) +(−ω^2 )^(107) = −ω^2 −ω = 1 .

x3+1=(x+1)(x2−x+1)rootsofx3+1=0are−1,−ω2,−ωSo(−ω)101+(−ω2)107=−ω2−ω=1.

Commented by Tawa1 last updated on 21/Oct/18

God bless you sir. but i don′t really understand sir. from the (− ω) and the substitution to get 1

Godblessyousir.butidon′treallyunderstandsir.fromthe(−ω)andthesubstitutiontoget1

Commented by $@ty@m last updated on 21/Oct/18

x^3 +1=0 ⇒(x+1)(x^2 −x+1)=0 ⇒x+1=0 or x^2 −x+1=0 ⇒x=−1 or x=−ω, −ω^2 i.e. roots of x^2 −x+1 are −ω, −ω^2 ∴ α=−ω, β=−ω^2

x3+1=0⇒(x+1)(x2−x+1)=0⇒x+1=0orx2−x+1=0⇒x=−1orx=−ω,−ω2i.e.rootsofx2−x+1are−ω,−ω2∴α=−ω,β=−ω2

Commented by Tawa1 last updated on 21/Oct/18

God bless you sir.

Godblessyousir.

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