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Question Number 128617 by ajfour last updated on 08/Jan/21

Commented by ajfour last updated on 08/Jan/21

Express x in terms of t, then use it to solve x^3 =x+c.

Expressxintermsoft,thenuseittosolvex3=x+c.

Answered by ajfour last updated on 09/Jan/21

x=p+(√(t^2 −q^2 )) Now x^3 =x+c ⇒ p^3 +(t^2 −q^2 )(√(t^2 −q^2 )) +3p(t^2 −q^2 )+3p^2 (√(t^2 −q^2 )) −p−(√(t^2 −q^2 ))=c Let 3p^2 −q^2 =1 ⇒ p^3 +t^2 (√(t^2 +1−3p^2 ))+3p(t^2 +1−3p^2 ) −p−c=0 ⇒ [8p^3 −(2+3t^2 )p+c]^2 =t^4 (t^2 +1−3p^2 ) let t^2 =mp^2 +s ⇒ [8p^3 −2p−3p(mp^2 +s)+c]^2 = (mp^2 +s)^2 (mp^2 +s+1−3p^2 ) ⇒ let s=−1 ⇒ 8p^3 −2p−3p(mp^2 −1)+c =(√(m−3))(mp^2 −1)p ⇒ (8−3m−m(√(m−3)))p^3 +(1+(√(m−3)))p+c = 0 let (1+(√(m−3)))=r r^3 =1+(m−3)(√(m−3)) +3(√(m−3))+3(m−3) = −8+m(√(m−3))+3m .....

x=p+t2−q2Nowx3=x+c⇒p3+(t2−q2)t2−q2+3p(t2−q2)+3p2t2−q2−p−t2−q2=cLet3p2−q2=1⇒p3+t2t2+1−3p2+3p(t2+1−3p2)−p−c=0⇒[8p3−(2+3t2)p+c]2=t4(t2+1−3p2)lett2=mp2+s⇒[8p3−2p−3p(mp2+s)+c]2=(mp2+s)2(mp2+s+1−3p2)⇒lets=−1⇒8p3−2p−3p(mp2−1)+c=m−3(mp2−1)p⇒(8−3m−mm−3)p3+(1+m−3)p+c=0let(1+m−3)=rr3=1+(m−3)m−3+3m−3+3(m−3)=−8+mm−3+3m.....

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