Menu Close

Question-159189

Tinku Tara 0 Comments
Pin




Question Number 159189 by mathlove last updated on 14/Nov/21
Answered by Rasheed.Sindhi last updated on 16/Nov/21
Let (A/B)=x x+(1/x)=1; x^(2019) +x^(−2019) =? ▶x+(1/x)=1 ⇒x^2 −x+1=0⇒(x+1)(x^2 −x+1)=0 ⇒x^3 +1=0⇒x^3 =−1 ∴ x is cuberoot of −1 ∴ x=−1,−ω,−ω^2 −1 is a root of x+1=0 −ω,−ω^2 are the roots of x^2 −x+1=0 where ω is cuberoot of unity. ▶x^(2019) +x^(−2019) =(−ω)^(2019) +(−ω)^(−2019) −(ω^(2019) +ω^(−2019) )=−({(ω)^3 }^(673) +{(ω)^3 }^(−673) ) =−(1^(673) +1^(673) )=−2 ((A/B))^(2019) +((B/A))^(2019) =−2
LetAB=xx+1x=1;x2019+x−2019=?▸x+1x=1⇒x2−x+1=0⇒(x+1)(x2−x+1)=0⇒x3+1=0⇒x3=−1∴xiscuberootof−1∴x=−1,−ω,−ω2−1isarootofx+1=0−ω,−ω2aretherootsofx2−x+1=0whereωiscuberootofunity.▸x2019+x−2019=(−ω)2019+(−ω)−2019−(ω2019+ω−2019)=−({(ω)3}673+{(ω)3}−673)=−(1673+1673)=−2(AB)2019+(BA)2019=−2

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *