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Question Number 90876 by john santu last updated on 26/Apr/20

((2cos ((π/9))))^(1/(3 )) −((2cos (((2π)/9))))^(1/(3 )) −((2cos (((4π)/9))))^(1/(3 )) = ?

2cos(π9)3−2cos(2π9)3−2cos(4π9)3=?

Commented by john santu last updated on 26/Apr/20

Ramanujan theorem let α, β ,γ be a roots x^3 −ax^2 +bx−1=0 then ((α ))^(1/(3 )) + (β)^(1/(3 )) + (γ)^(1/(3 )) = ((a+6+3t))^(1/(3 )) with t^3 −3(a+b+3)t−(ab+6(a+b)+9)=0 ⇒ (α)^(1/(3 )) +(β)^(1/(3 )) +(γ)^(1/(3 )) = ((a+6+3t))^(1/(3 )) = ϕ ((αβ))^(1/(3 )) + ((βγ))^(1/(3 )) + ((αγ))^(1/(3 )) = ((b+6+3t))^(1/(3 )) = θ (t+3)^3 = (ϕθ)^3 consider x^3 −3x−1=0 x = t+(1/t) ⇒t^9 +1 = 0 t^9 = −1 = e^(iπ+2πki) = e^(iπ(1+2k)) ,k∈Z t = e^(iπ(((1+2k)/9))) , k = 0,1,2,...,8 t = e^((iπ)/9) , e^((i5π)/9) , e^((i7π)/9) t = e^(iθ) = cos θ+isin θ x^3 −3x−1 = 0 roots α = 2cos ((π/9)) β= −2cos (((2π)/9)) γ = −2cos (((4π)/9)) ∴ ((2cos ((π/9))))^(1/(3 )) −((2cos (((2π)/9))))^(1/(3 )) −((2cos (((4π)/9))))^(1/(3 )) = ((6−3(9)^(1/(3 )) ))^(1/(3 ))

Ramanujantheoremletα,β,γbearootsx3−ax2+bx−1=0thenα3+β3+γ3=a+6+3t3witht3−3(a+b+3)t−(ab+6(a+b)+9)=0⇒α3+β3+γ3=a+6+3t3=φαβ3+βγ3+αγ3=b+6+3t3=θ(t+3)3=(φθ)3considerx3−3x−1=0x=t+1t⇒t9+1=0t9=−1=eiπ+2πki=eiπ(1+2k),k∈Zt=eiπ(1+2k9),k=0,1,2,...,8t=eiπ9,ei5π9,ei7π9t=eiθ=cosθ+isinθx3−3x−1=0rootsα=2cos(π9)β=−2cos(2π9)γ=−2cos(4π9)∴2cos(π9)3−2cos(2π9)3−2cos(4π9)3=6−3933

Commented by jagoll last updated on 26/Apr/20

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