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Question Number 164341 by ajfour last updated on 16/Jan/22
x^4 =x^2 +cx (x^2 −(1/2))^2 =cx+(1/4) let x^2 −(1/2)=t ⇒ (t^2 −(1/4))^2 =c^2 (t+(1/2)) now let t^2 −(1/4)=z ⇒ (z^2 −(c^2 /2))^2 =c^4 (z+(1/4)) z^2 −(c^2 /2)=p ⇒ (p^2 −(c^4 /4))^2 =c^8 (p+(c^2 /2)) ⇒ p^4 −((c^4 p^2 )/2)−c^8 p+(c^8 /(16))−(c^(10) /2)=0 p^4 −Ap^2 −Bp−λ=0 (p^2 +sp+h)(p^2 −sp−(λ/h))=0 h−(λ/h)=s^2 −A s(h+(λ/h))=B (B^2 /s^2 )−(s^2 −A)^2 =4λ (s^2 −A)^3 +A(s^2 −A)^2 +4λ(s^2 −A) +4λA−B^2 =0 m^3 +Am^2 +4λm+(4λA−B^2 )=0 let m=w−(A/3) ⇒ w^3 +(4λ−(A^2 /3))w+((2A^3 )/(27))+((8λA)/3)−B^2 =0 4λ−(A^2 /3)=2c^(10) −(c^8 /4)−(c^8 /(12)) =2c^8 (c^2 −(1/6))
$$\:\:{x}^{\mathrm{4}} ={x}^{\mathrm{2}} +{cx} \\ $$$$\left({x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} ={cx}+\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${let}\:\:\:\:{x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}={t}\:\:\:\Rightarrow \\ $$$$\left({t}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} ={c}^{\mathrm{2}} \left({t}+\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$${now}\:\:{let}\:\:\:{t}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}={z}\:\:\:\Rightarrow \\ $$$$\left({z}^{\mathrm{2}} −\frac{{c}^{\mathrm{2}} }{\mathrm{2}}\right)^{\mathrm{2}} ={c}^{\mathrm{4}} \left({z}+\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$$${z}^{\mathrm{2}} −\frac{{c}^{\mathrm{2}} }{\mathrm{2}}={p}\:\:\Rightarrow\:\:\left({p}^{\mathrm{2}} −\frac{{c}^{\mathrm{4}} }{\mathrm{4}}\right)^{\mathrm{2}} ={c}^{\mathrm{8}} \left({p}+\frac{{c}^{\mathrm{2}} }{\mathrm{2}}\right) \\ $$$$\Rightarrow\:\:\:{p}^{\mathrm{4}} −\frac{{c}^{\mathrm{4}} {p}^{\mathrm{2}} }{\mathrm{2}}−{c}^{\mathrm{8}} {p}+\frac{{c}^{\mathrm{8}} }{\mathrm{16}}−\frac{{c}^{\mathrm{10}} }{\mathrm{2}}=\mathrm{0} \\ $$$${p}^{\mathrm{4}} −{Ap}^{\mathrm{2}} −{Bp}−\lambda=\mathrm{0} \\ $$$$\left({p}^{\mathrm{2}} +{sp}+{h}\right)\left({p}^{\mathrm{2}} −{sp}−\frac{\lambda}{{h}}\right)=\mathrm{0} \\ $$$${h}−\frac{\lambda}{{h}}={s}^{\mathrm{2}} −{A} \\ $$$${s}\left({h}+\frac{\lambda}{{h}}\right)={B} \\ $$$$\frac{{B}^{\mathrm{2}} }{{s}^{\mathrm{2}} }−\left({s}^{\mathrm{2}} −{A}\right)^{\mathrm{2}} =\mathrm{4}\lambda \\ $$$$\left({s}^{\mathrm{2}} −{A}\right)^{\mathrm{3}} +{A}\left({s}^{\mathrm{2}} −{A}\right)^{\mathrm{2}} +\mathrm{4}\lambda\left({s}^{\mathrm{2}} −{A}\right) \\ $$$$\:\:\:\:\:\:\:\:\:+\mathrm{4}\lambda{A}−{B}^{\mathrm{2}} =\mathrm{0} \\ $$$${m}^{\mathrm{3}} +{Am}^{\mathrm{2}} +\mathrm{4}\lambda{m}+\left(\mathrm{4}\lambda{A}−{B}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$${let}\:\:\:{m}={w}−\frac{{A}}{\mathrm{3}}\:\:\:\Rightarrow \\ $$$${w}^{\mathrm{3}} +\left(\mathrm{4}\lambda−\frac{{A}^{\mathrm{2}} }{\mathrm{3}}\right){w}+\frac{\mathrm{2}{A}^{\mathrm{3}} }{\mathrm{27}}+\frac{\mathrm{8}\lambda{A}}{\mathrm{3}}−{B}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{4}\lambda−\frac{{A}^{\mathrm{2}} }{\mathrm{3}}=\mathrm{2}{c}^{\mathrm{10}} −\frac{{c}^{\mathrm{8}} }{\mathrm{4}}−\frac{{c}^{\mathrm{8}} }{\mathrm{12}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}{c}^{\mathrm{8}} \left({c}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{6}}\right) \\ $$
Commented by Tawa11 last updated on 16/Jan/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$

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