x-2-x-2-1-1-c-x-3-c-x-3- Tinku Tara June 4, 2023 Algebra 0 Comments FacebookTweetPin Question Number 192914 by ajfour last updated on 30/May/23 x2(x2−1)=(1−cx)3+(cx)3 Answered by a.lgnaoui last updated on 31/May/23 x2(x2−1)=(x−c)3+c3x3x5(x2−1)=x3−3cx(x−c)x4(x2−1)=x2−3c(x−c)x6−x4=(x−3c2)2+3c24[(x−3c2)+3c2]6=[(x−3c2)+3c2]4+(x−3c2)2+3c24X=x−aavec3c2=a(X+a)6=(X+a)4+X2+ac2(X+a)6−(X+a)4−[(X+a)−a]2+ac2X+a=ZZ6−Z4−(Z2−2aZ+a2)−ac2=0Z6−Z4−Z2+2aZ−a(2a+c)2=0soitZ6−Z4−Z2+3cZ−3c2=0Z=X+a=xx6−x4−x2+3c(x−c)=0……….. Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: consider-the-triple-of-real-numbers-x-y-z-defined-by-the-addittion-x-y-z-x-y-z-x-x-y-y-z-z-and-scalar-multiplication-by-x-y-z-0-0-0-Show-that-all-axioms-for-a-vector-space-are-Next Next post: consider-the-space-Pn-with-H-f-f-Pn-and-0-1-f-x-x-0-Show-that-H-is-a-SUBSPACE-of-Pn- Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment * Name * Save my name, email, and website in this browser for the next time I comment.
![x^2 (x^2 −1)=(((x−c)^3 +c^3 )/x^3 ) x^5 (x^2 −1)=x^3 −3cx(x−c) x^4 (x^2 −1)=x^2 −3c(x−c) x^6 −x^4 = ( x−((3c)/2))^2 +((3c^2 )/4) [(x−((3c)/2))+((3c)/2)]^6 =[(x−((3c)/2))+((3c)/2)]^4 +(x−((3c)/2))^2 +((3c^2 )/4) X=x−a avec ((3c)/2)=a (X+a)^6 =(X+a)^4 +X^2 +((ac)/2) (X+a)^6 −(X+a)^4 −[(X+a)−a]^2 +((ac)/2) X+a=Z Z^6 −Z^4 −(Z^2 −2aZ+a^2 )−((ac)/2)=0 Z^6 −Z^4 −Z^2 +2aZ−((a(2a+c))/2) =0 soit Z^6 −Z^4 −Z^2 +3cZ−3c^2 =0 Z=X+a=x x^6 −x^4 −x^2 +3c(x−c)=0 ...........](https://www.tinkutara.com/question/Q192940.png)