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a-b-1-b-1-i-b-1-b-a-2-ii-Solve-simultaneously-for-a-and-b-

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Question Number 36221 by ajfour last updated on 30/May/18
a(b+(1/b))=−1 ....(i) b−(1/b)=a^2 ....(ii) Solve simultaneously for a, and b.
a(b+1b)=−1….(i)b−1b=a2….(ii)Solvesimultaneouslyfora,andb.
Commented by behi83417@gmail.com last updated on 30/May/18
a^2 (b^2 +b^(−2) +2)=1⇒(b−b^(−1) )(b^2 +b^(−2) +2)=1 b^3 +b^(−1) +2b−b−b^(−3) −2b^(−1) =1 b^3 +b−b^(−1) −b^(−3) =1⇒b^6 +b^4 −b^2 −1=b^3 ⇒b^6 +b^4 −b^3 −b^2 −1=0....
a2(b2+b−2+2)=1⇒(b−b−1)(b2+b−2+2)=1b3+b−1+2b−b−b−3−2b−1=1b3+b−b−1−b−3=1⇒b6+b4−b2−1=b3⇒b6+b4−b3−b2−1=0….
Answered by Rasheed.Sindhi last updated on 30/May/18
a(b+(1/b))=−1 ....(i) b−(1/b)=a^2 ....(ii) (i)⇒b+(1/b)=−1/a (b+(1/b))^2 =(−1/a)^2 ⇒b^2 +(1/b^2 )=(1/a^2 )−2..........(iii) (ii)⇒(b−(1/b))^2 =(a^2 )^2 ⇒b^2 +(1/b^2 )=a^4 +2............(iv) (iii)&(iv): a^4 +2=(1/a^2 )−2 a^6 +4a^2 −1=0 (i)⇒a(b+(1/b))^2 =(1/a^2 ) ...........(v) (ii)×(v): (b−(1/b))(b+(1/b))^2 =1 (b^2 −(1/b^2 ))(b+(1/b))=1 b^3 −(1/b^3 )+b−(1/b)=1 (b−(1/b))^3 +4(b−(1/b))=1 Continue
a(b+1b)=−1….(i)b−1b=a2….(ii)(i)⇒b+1b=−1/a(b+1b)2=(−1/a)2⇒b2+1b2=1a2−2……….(iii)(ii)⇒(b−1b)2=(a2)2⇒b2+1b2=a4+2…………(iv)(iii)&(iv):a4+2=1a2−2a6+4a2−1=0(i)⇒a(b+1b)2=1a2………..(v)(ii)×(v):(b−1b)(b+1b)2=1(b2−1b2)(b+1b)=1b3−1b3+b−1b=1(b−1b)3+4(b−1b)=1Continue
Commented by MJS last updated on 30/May/18
a^6 +4a^2 −1=0 a=(√t) t^3 +4t−1=0 solution almost the same as of the former problem (w^3 +4w+1=0) t=((−(q/2)+(√((p^3 /(27))+(q^2 /4)))))^(1/3) +((−(q/2)−(√((p^3 /(27))+(q^2 /4)))))^(1/3) = =(((1/2)+((√(849))/(18))))^(1/3) +(((1/2)−((√(849))/(18))))^(1/3)
a6+4a2−1=0a=tt3+4t−1=0solutionalmostthesameasoftheformerproblem(w3+4w+1=0)t=−q2+p327+q243+−q2−p327+q243==12+849183+12−849183
Commented by ajfour last updated on 30/May/18
thank you sir!
thankyousir!
Answered by tanmay.chaudhury50@gmail.com last updated on 30/May/18
a^2 (b^2 +2+(1/b^2 ))=1 (b−(1/b))(b^2 +2+(1/b^2 ))=1 (((b^2 −1)/b))(((b^4 +2b^2 +1)/b^2 ))=1 b^6 +2b^4 +b^2 −b^4 −2b^2 −1=b^3 b^6 +b^4 −b^3 −b^2 −1=0 b^3 +b−1−(1/b)−(1/b^3 )=0 (b^3 −(1/b^3 ))+(b−(1/b))−1=0 b−(1/b)=k (b−(1/b))^3 =k^3 b^3 −3b^2 .(1/b)+3b.(1/b^2 )−(1/b^3 )=k^3 b^3 −(1/b^3 )−3(b−(1/b))=k^3 so b^3 −(1/b^3 )=k^3 +3k so eauation is k^3 +3k+k−1=0 k^3 +4k−1=0 k=(z/n) ((z/n))^3 +4((z/n))−1=0 z^3 +4zn^2 −n^3 =0 cos3θ=4cos^3 θ−3cosθ cos^3 θ−(1/4)cos3θ−(3/4)cosθ=0
a2(b2+2+1b2)=1(b−1b)(b2+2+1b2)=1(b2−1b)(b4+2b2+1b2)=1b6+2b4+b2−b4−2b2−1=b3b6+b4−b3−b2−1=0b3+b−1−1b−1b3=0(b3−1b3)+(b−1b)−1=0b−1b=k(b−1b)3=k3b3−3b2.1b+3b.1b2−1b3=k3b3−1b3−3(b−1b)=k3sob3−1b3=k3+3ksoeauationisk3+3k+k−1=0k3+4k−1=0k=zn(zn)3+4(zn)−1=0z3+4zn2−n3=0cos3θ=4cos3θ−3cosθcos3θ−14cos3θ−34cosθ=0
Commented by tanmay.chaudhury50@gmail.com last updated on 30/May/18
Commented by tanmay.chaudhury50@gmail.com last updated on 30/May/18

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