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))


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Question Number 164341 by ajfour last updated on 16/Jan/22

$x^{4} = x^{2} + c x$ ${(x^{2} - \frac{1}{2})}^{2} = c x + \frac{1}{4}$ $l e t x^{2} - \frac{1}{2} = t \Rightarrow$ ${(t^{2} - \frac{1}{4})}^{2} = c^{2} (t + \frac{1}{2})$ $n o w l e t t^{2} - \frac{1}{4} = z \Rightarrow$ ${(z^{2} - \frac{c^{2}}{2})}^{2} = c^{4} (z + \frac{1}{4})$ $z^{2} - \frac{c^{2}}{2} = p \Rightarrow {(p^{2} - \frac{c^{4}}{4})}^{2} = c^{8} (p + \frac{c^{2}}{2})$ $\Rightarrow p^{4} - \frac{c^{4} p^{2}}{2} - c^{8} p + \frac{c^{8}}{16} - \frac{c^{10}}{2} = 0$ $p^{4} - {A p}^{2} - B p - λ = 0$ $(p^{2} + s p + h) (p^{2} - s p - \frac{λ}{h}) = 0$ $h - \frac{λ}{h} = s^{2} - A$ $s (h + \frac{λ}{h}) = B$ $\frac{B^{2}}{s^{2}} - {(s^{2} - A)}^{2} = 4 λ$ ${(s^{2} - A)}^{3} + A {(s^{2} - A)}^{2} + 4 λ (s^{2} - A)$ $+ 4 λ A - B^{2} = 0$ $m^{3} + {A m}^{2} + 4 λ m + (4 λ A - B^{2}) = 0$ $l e t m = w - \frac{A}{3} \Rightarrow$ $w^{3} + (4 λ - \frac{A^{2}}{3}) w + \frac{2 A^{3}}{27} + \frac{8 λ A}{3} - B^{2} = 0$ $4 λ - \frac{A^{2}}{3} = 2 c^{10} - \frac{c^{8}}{4} - \frac{c^{8}}{12}$ $= 2 c^{8} (c^{2} - \frac{1}{6})$
Commented by Tawa11 last updated on 16/Jan/22

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