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Question Number 128090 by ajfour last updated on 04/Jan/21
someone help cheking this: x^3 =x+c let x=(p−2q)+(q−2p) = −(p+q) p^3 −8q^3 −6pq(p−2q)+ q^3 −8p^3 −6pq(q−2p)+ 3(p−2q)(q−2p)[(p−2q)+(q−2p)] = (p−2q)+(q−2p)+c ⇒ −7(p^3 +q^3 )+6pq(p+q) −3(p+q)[5pq−2(p^2 +q^2 )] +(p+q) = c let p^3 +q^3 =−(c/7) And since x=−(p+q) so with −x^3 =−x−c we have p^3 +q^3 +3pq(p+q)=p+q−c ⇒ (3pq−1)(p+q)=−((6c)/7) 1−9pq+6(p^2 +q^2 )=0 (p^2 +q^2 )(p+q)−pq(p+q)=−(c/7) say p^2 +q^2 =t , then pq=((1+6t)/9) ; p+q=(((−((6c)/7)))/(((1+6t)/3)−1)) p+q=((9c)/(7(1−3t))) p^2 +q^2 =(p+q)^2 −2pq t=[((9c)/(7(1−3t)))]^2 −((2(1+6t))/9) [((9c)/(7(1−3t)))]^2 =((21t+2)/9) ⇒ (21t+2)(3t−1)^2 =(((27c)/7))^2 .....
someonehelpchekingthis:x3=x+cletx=(p−2q)+(q−2p)=−(p+q)p3−8q3−6pq(p−2q)+q3−8p3−6pq(q−2p)+3(p−2q)(q−2p)[(p−2q)+(q−2p)]=(p−2q)+(q−2p)+c⇒−7(p3+q3)+6pq(p+q)−3(p+q)[5pq−2(p2+q2)]+(p+q)=cletp3+q3=−c7Andsincex=−(p+q)sowith−x3=−x−cwehavep3+q3+3pq(p+q)=p+q−c⇒(3pq−1)(p+q)=−6c71−9pq+6(p2+q2)=0(p2+q2)(p+q)−pq(p+q)=−c7sayp2+q2=t,thenpq=1+6t9;p+q=(−6c7)1+6t3−1p+q=9c7(1−3t)p2+q2=(p+q)2−2pqt=[9c7(1−3t)]2−2(1+6t)9[9c7(1−3t)]2=21t+29⇒(21t+2)(3t−1)2=(27c7)2…..
Commented by MJS_new last updated on 04/Jan/21
after setting 6pq=−1 ⇒ q=−(1/(6p)) and you have a fixed value after −7(p^3 +q^3 )=c because c is given
aftersetting6pq=−1⇒q=−16pandyouhaveafixedvalueafter−7(p3+q3)=cbecausecisgiven
Commented by MJS_new last updated on 04/Jan/21
you′re right in this point still I don′t get your equation after substituting x=(p−2q)+(q−2p)
you′rerightinthispointstillIdon′tgetyourequationaftersubstitutingx=(p−2q)+(q−2p)

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