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.