x^3 =x+c x^4 =x^2 +cx let cx=kx^4 +hx^2 ⇒ kx^3 +hx=c x^2 =1+c(kx^2 +h) x^2 =((1+ch)/(1−ck)) (((1+ch)/(1−ck))){((k(1+ch))/(1−ck))+h}^2 =c^2 (1+ch)(h+k)^2 =c^2 (1−ck)^3 ⇒ (1+ch)(h^2 +k^2 +2hk) =c^2 (1−c^3 k^3 +3c^2 k^2 −3ck) c^5 k^3 +(1+ch−3c^4 )k^2 +{2h(1+ch)+3c^3 }k +(1+ch)h^2 −c^2 =0 let h=3c^3 −(1/c) ⇒ c^5 k^3 +{9c^3 −(2/c)+2c(3c^3 −(1/c))^2 }k +{3c^4 (3c^3 −(1/c))^2 −c^2 }=0 ⇒ k^3 +3(6c^2 −(1/c^2 ))k +27c^5 −18c+(2/c^3 )=0 D=(((27c^5 )/2)+9c+(1/c^3 ))^2 +(6c^2 −(1/c^2 ))^3 ...


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Question Number 156639 by ajfour last updated on 13/Oct/21
$x^3 =x+c x^4 =x^2 +cx let cx=kx^4 +hx^2 ⇒ kx^3 +hx=c x^2 =1+c(kx^2 +h) x^2 =((1+ch)/(1−ck)) (((1+ch)/(1−ck))){((k(1+ch))/(1−ck))+h}^2 =c^2 (1+ch)(h+k)^2 =c^2 (1−ck)^3 ⇒ (1+ch)(h^2 +k^2 +2hk) =c^2 (1−c^3 k^3 +3c^2 k^2 −3ck) c^5 k^3 +(1+ch−3c^4 )k^2 +{2h(1+ch)+3c^3 }k +(1+ch)h^2 −c^2 =0 let h=3c^3 −(1/c) ⇒ c^5 k^3 +{9c^3 −(2/c)+2c(3c^3 −(1/c))^2 }k +{3c^4 (3c^3 −(1/c))^2 −c^2 }=0 ⇒ k^3 +3(6c^2 −(1/c^2 ))k +27c^5 −18c+(2/c^3 )=0 D=(((27c^5 )/2)+9c+(1/c^3 ))^2 +(6c^2 −(1/c^2 ))^3 ...$
$x^{3} = x + c$ $x^{4} = x^{2} + cx$ $let cx = {kx}^{4} + {hx}^{2}$ $\Rightarrow {kx}^{3} + hx = c$ $x^{2} = 1 + c ({kx}^{2} + h)$ $x^{2} = \frac{1 + ch}{1 - ck}$ $(\frac{1 + ch}{1 - ck}) {\frac{k (1 + ch)}{1 - ck} + h}^{2} = c^{2}$ $(1 + ch) {(h + k)}^{2} = c^{2} {(1 - ck)}^{3}$ $\Rightarrow (1 + ch) (h^{2} + k^{2} + 2 hk)$ $= c^{2} (1 - c^{3} k^{3} + {3 c}^{2} k^{2} - 3 ck)$ $c^{5} k^{3} + (1 + ch - {3 c}^{4}) k^{2}$ $+ {2 h (1 + ch) + {3 c}^{3}} k$ $+ (1 + ch) h^{2} - c^{2} = 0$ $let h = {3 c}^{3} - \frac{1}{c} \Rightarrow$ $c^{5} k^{3} + {{9 c}^{3} - \frac{2}{c} + 2 c {({3 c}^{3} - \frac{1}{c})}^{2}} k$ $+ {{3 c}^{4} {({3 c}^{3} - \frac{1}{c})}^{2} - c^{2}} = 0$ $\Rightarrow k^{3} + 3 ({6 c}^{2} - \frac{1}{c^{2}}) k$ $+ {27 c}^{5} - 18 c + \frac{2}{c^{3}} = 0$ $D = {(\frac{{27 c}^{5}}{2} + 9 c + \frac{1}{c^{3}})}^{2} + {({6 c}^{2} - \frac{1}{c^{2}})}^{3}$ $. . .$
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