Question Number 100492 by bobhans last updated on 27/Jun/20

((16−((64)/(16−((64)/(16−((64)/(16−...))))))))^(1/(3 )) −((−2−(1/(−2−(1/(−2−(1/(−2−...))))))))^(1/(3 ))

Commented byDwaipayan Shikari last updated on 27/Jun/20

p=16−((64)/(16−((64)/(16−((64)/(.....)))))) p=16−((64)/p)⇒p^2 −16p+64=0⇒p=8 y=−2−(1/(−2−(1/(−2−(1/(.....)))))) y=−2−(1/y)⇒y^2 +2y+1=0⇒ y=−1 (p)^(1/3) −(y)^(1/3) =2−((−1))^(1/3) (imaginary value for second sum) Or 2−(−1)=3 But it also takes two complex roots

Commented byfloor(10²Eta[1]) last updated on 27/Jun/20

bro the imaginary value is (√(−1)) the ^3 (√(−1)) it′s just −1 [(−1)^3 =−1] (and the other 2 complex roots)

Commented byCoronavirus last updated on 27/Jun/20

🙇🙇🙇

Answered by bramlex last updated on 27/Jun/20

let q = ((16−((64)/(16−((64)/(16−...))))))^(1/3) q^3 = 16−((64)/q) ⇒ q^4 −16q+64 = 0 let r = ((−2−(1/(−2−(1/(−2−...))))))^(1/(3 )) r^3 = −2−(1/r) ⇒ r^4 +2r+1=0 ; r=−1

Answered by floor(10²Eta[1]) last updated on 27/Jun/20

let x=16−((64)/(16−((64)/(16−...)))) and y=−2−(1/(−2−(1/(−2−...)))) so we want to know ^3 (√x)−^3 (√y). x=16−((64)/x)⇒x^2 −16x+64=0⇒x=8 y=−2−(1/y)⇒y^2 +2y+1=0⇒y=−1 so ^3 (√x)−^3 (√y)=^3 (√8)−^3 (√(−1))=2−(−1)=3.