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

Question Number 36221 by ajfour last updated on 30/May/18

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....

$${a}^{\mathrm{2}} \left({b}^{\mathrm{2}} +{b}^{−\mathrm{2}} +\mathrm{2}\right)=\mathrm{1}\Rightarrow\left({b}−{b}^{−\mathrm{1}} \right)\left({b}^{\mathrm{2}} +{b}^{−\mathrm{2}} +\mathrm{2}\right)=\mathrm{1} \\ $$$${b}^{\mathrm{3}} +{b}^{−\mathrm{1}} +\mathrm{2}{b}−{b}−{b}^{−\mathrm{3}} −\mathrm{2}{b}^{−\mathrm{1}} =\mathrm{1} \\ $$$${b}^{\mathrm{3}} +{b}−{b}^{−\mathrm{1}} −{b}^{−\mathrm{3}} =\mathrm{1}\Rightarrow{b}^{\mathrm{6}} +{b}^{\mathrm{4}} −{b}^{\mathrm{2}} −\mathrm{1}={b}^{\mathrm{3}} \\ $$$$\Rightarrow{b}^{\mathrm{6}} +{b}^{\mathrm{4}} −{b}^{\mathrm{3}} −{b}^{\mathrm{2}} −\mathrm{1}=\mathrm{0}…. \\ $$$$ \\ $$

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

$${a}\left({b}+\frac{\mathrm{1}}{{b}}\right)=−\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:….\left({i}\right) \\ $$$${b}−\frac{\mathrm{1}}{{b}}={a}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:….\left({ii}\right) \\ $$$$\left(\mathrm{i}\right)\Rightarrow{b}+\frac{\mathrm{1}}{{b}}=−\mathrm{1}/{a} \\ $$$$\:\:\:\:\:\:\:\:\:\left({b}+\frac{\mathrm{1}}{{b}}\right)^{\mathrm{2}} =\left(−\mathrm{1}/{a}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\Rightarrow{b}^{\mathrm{2}} +\frac{\mathrm{1}}{{b}^{\mathrm{2}} }=\frac{\mathrm{1}}{{a}^{\mathrm{2}} }−\mathrm{2}……….\left(\mathrm{iii}\right) \\ $$$$\left(\mathrm{ii}\right)\Rightarrow\left({b}−\frac{\mathrm{1}}{{b}}\right)^{\mathrm{2}} =\left({a}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\Rightarrow{b}^{\mathrm{2}} +\frac{\mathrm{1}}{{b}^{\mathrm{2}} }={a}^{\mathrm{4}} +\mathrm{2}…………\left(\mathrm{iv}\right) \\ $$$$\left(\mathrm{iii}\right)\&\left(\mathrm{iv}\right):\:{a}^{\mathrm{4}} +\mathrm{2}=\frac{\mathrm{1}}{{a}^{\mathrm{2}} }−\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{a}^{\mathrm{6}} +\mathrm{4}{a}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$$\left(\mathrm{i}\right)\Rightarrow{a}\left({b}+\frac{\mathrm{1}}{{b}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{{a}^{\mathrm{2}} }\:………..\left(\mathrm{v}\right)\:\:\:\:\:\:\:\:\: \\ $$$$\left(\mathrm{ii}\right)×\left(\mathrm{v}\right):\:\:\: \\ $$$$\left({b}−\frac{\mathrm{1}}{{b}}\right)\left({b}+\frac{\mathrm{1}}{{b}}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$$\left({b}^{\mathrm{2}} −\frac{\mathrm{1}}{{b}^{\mathrm{2}} }\right)\left({b}+\frac{\mathrm{1}}{{b}}\right)=\mathrm{1} \\ $$$${b}^{\mathrm{3}} −\frac{\mathrm{1}}{{b}^{\mathrm{3}} }+{b}−\frac{\mathrm{1}}{{b}}=\mathrm{1} \\ $$$$\left({b}−\frac{\mathrm{1}}{{b}}\right)^{\mathrm{3}} +\mathrm{4}\left({b}−\frac{\mathrm{1}}{{b}}\right)=\mathrm{1} \\ $$$${Continue} \\ $$

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)

$${a}^{\mathrm{6}} +\mathrm{4}{a}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$${a}=\sqrt{{t}} \\ $$$${t}^{\mathrm{3}} +\mathrm{4}{t}−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{solution}\:\mathrm{almost}\:\mathrm{the}\:\mathrm{same}\:\mathrm{as}\:\mathrm{of}\:\mathrm{the}\:\mathrm{former} \\ $$$$\mathrm{problem}\:\left({w}^{\mathrm{3}} +\mathrm{4}{w}+\mathrm{1}=\mathrm{0}\right) \\ $$$$ \\ $$$${t}=\sqrt[{\mathrm{3}}]{−\frac{{q}}{\mathrm{2}}+\sqrt{\frac{{p}^{\mathrm{3}} }{\mathrm{27}}+\frac{{q}^{\mathrm{2}} }{\mathrm{4}}}}+\sqrt[{\mathrm{3}}]{−\frac{{q}}{\mathrm{2}}−\sqrt{\frac{{p}^{\mathrm{3}} }{\mathrm{27}}+\frac{{q}^{\mathrm{2}} }{\mathrm{4}}}}= \\ $$$$=\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{849}}}{\mathrm{18}}}+\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{849}}}{\mathrm{18}}} \\ $$

Commented by ajfour last updated on 30/May/18

$${thank}\:{you}\:{sir}! \\ $$

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

$${a}^{\mathrm{2}} \left({b}^{\mathrm{2}} +\mathrm{2}+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }\right)=\mathrm{1} \\ $$$$\left({b}−\frac{\mathrm{1}}{{b}}\right)\left({b}^{\mathrm{2}} +\mathrm{2}+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }\right)=\mathrm{1} \\ $$$$\left(\frac{{b}^{\mathrm{2}} −\mathrm{1}}{{b}}\right)\left(\frac{{b}^{\mathrm{4}} +\mathrm{2}{b}^{\mathrm{2}} +\mathrm{1}}{{b}^{\mathrm{2}} }\right)=\mathrm{1} \\ $$$${b}^{\mathrm{6}} +\mathrm{2}{b}^{\mathrm{4}} +{b}^{\mathrm{2}} −{b}^{\mathrm{4}} −\mathrm{2}{b}^{\mathrm{2}} −\mathrm{1}={b}^{\mathrm{3}} \\ $$$${b}^{\mathrm{6}} +{b}^{\mathrm{4}} −{b}^{\mathrm{3}} −{b}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$${b}^{\mathrm{3}} +{b}−\mathrm{1}−\frac{\mathrm{1}}{{b}}−\frac{\mathrm{1}}{{b}^{\mathrm{3}} }=\mathrm{0} \\ $$$$\left({b}^{\mathrm{3}} −\frac{\mathrm{1}}{{b}^{\mathrm{3}} }\right)+\left({b}−\frac{\mathrm{1}}{{b}}\right)−\mathrm{1}=\mathrm{0} \\ $$$${b}−\frac{\mathrm{1}}{{b}}={k} \\ $$$$\left({b}−\frac{\mathrm{1}}{{b}}\right)^{\mathrm{3}} ={k}^{\mathrm{3}} \\ $$$${b}^{\mathrm{3}} −\mathrm{3}{b}^{\mathrm{2}} .\frac{\mathrm{1}}{{b}}+\mathrm{3}{b}.\frac{\mathrm{1}}{{b}^{\mathrm{2}} }−\frac{\mathrm{1}}{{b}^{\mathrm{3}} }={k}^{\mathrm{3}} \\ $$$${b}^{\mathrm{3}} −\frac{\mathrm{1}}{{b}^{\mathrm{3}} }−\mathrm{3}\left({b}−\frac{\mathrm{1}}{{b}}\right)={k}^{\mathrm{3}} \\ $$$${so}\:{b}^{\mathrm{3}} −\frac{\mathrm{1}}{{b}^{\mathrm{3}} }={k}^{\mathrm{3}} +\mathrm{3}{k} \\ $$$${so}\:{eauation}\:{is} \\ $$$${k}^{\mathrm{3}} +\mathrm{3}{k}+{k}−\mathrm{1}=\mathrm{0} \\ $$$${k}^{\mathrm{3}} +\mathrm{4}{k}−\mathrm{1}=\mathrm{0} \\ $$$${k}=\frac{{z}}{{n}} \\ $$$$\left(\frac{{z}}{{n}}\right)^{\mathrm{3}} +\mathrm{4}\left(\frac{{z}}{{n}}\right)−\mathrm{1}=\mathrm{0} \\ $$$${z}^{\mathrm{3}} +\mathrm{4}{zn}^{\mathrm{2}} −{n}^{\mathrm{3}} =\mathrm{0} \\ $$$${cos}\mathrm{3}\theta=\mathrm{4}{cos}^{\mathrm{3}} \theta−\mathrm{3}{cos}\theta \\ $$$${cos}^{\mathrm{3}} \theta−\frac{\mathrm{1}}{\mathrm{4}}{cos}\mathrm{3}\theta−\frac{\mathrm{3}}{\mathrm{4}}{cos}\theta=\mathrm{0} \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 30/May/18