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if-x-5-x-4-x-3-2x-2-x-1-0-find-x-3-1-x-3-

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Question Number 153130 by mathdanisur last updated on 04/Sep/21
if x^5 +x^4 +x^3 +2x^2 +x+1=0 find x^3 - (1/x^3 ) = ?
ifx5+x4+x3+2x2+x+1=0findx3−1x3=?
Answered by ajfour last updated on 05/Sep/21
((x^6 −1)/(x−1))+x^2 =0 x^6 +x^3 −x^2 −1=0 (x^6 +x^3 −1)^3 =x^6 x^3 −(1/x^3 )=t ⇒ x^6 =tx^3 +1 now (t+1)^3 x^9 =tx^3 +1 (t+1)^3 (tx^3 +1)x^3 =tx^3 +1 ⇒ t(t+1)^3 (tx^3 +1)+x^3 =1 ⇒ x^3 =((1−t(t+1)^3 )/(1+t^2 (t+1)^3 )) ((1−t(t+1)^3 )/(1+t^2 (t+1)^3 ))−((1+t^2 (t+1)^3 )/(1−t(t+1)^3 ))=t ⇒ 1+t^2 (t+1)^6 −2t(t+1)^3 −1−t^4 (t+1)^6 −2t^2 (t+1)^3 =t−t^2 (t+1)^3 +t^3 (t+1)^3 −t^4 (t+1)^6 ⇒ t(t+1)^6 −(2+t+t^2 )(t+1)^3 =1 ⇒(t+1)^3 {t(t+1)^3 −t^2 −t−2}=1 ⇒ p^3 (p^3 t−pt−2)=1 & p−t=1 ⇒ p^3 {t^2 (p+1)−2}=1 p^3 {(p+1)^3 −2}=1 p^3 q^3 =1+2p^3 q=p+1 p^3 (q^3 −p^3 )=1+p^3 p^3 (p^2 +q^2 +pq)=q(p^2 −p+1) p^3 (1+3pq)=q(p^2 −p+1) p^3 (3p^2 +3p+1)=1+p^3 3p^5 +3p^4 =1 p^4 (p+1)=(1/3) ⇒ 3(t+1)^4 t=1 now use any polynomial calculator, say,wolframalpha.
x6−1x−1+x2=0x6+x3−x2−1=0(x6+x3−1)3=x6x3−1x3=t⇒x6=tx3+1now(t+1)3x9=tx3+1(t+1)3(tx3+1)x3=tx3+1⇒t(t+1)3(tx3+1)+x3=1⇒x3=1−t(t+1)31+t2(t+1)31−t(t+1)31+t2(t+1)3−1+t2(t+1)31−t(t+1)3=t⇒1+t2(t+1)6−2t(t+1)3−1−t4(t+1)6−2t2(t+1)3=t−t2(t+1)3+t3(t+1)3−t4(t+1)6⇒t(t+1)6−(2+t+t2)(t+1)3=1⇒(t+1)3{t(t+1)3−t2−t−2}=1⇒p3(p3t−pt−2)=1&p−t=1⇒p3{t2(p+1)−2}=1p3{(p+1)3−2}=1p3q3=1+2p3q=p+1p3(q3−p3)=1+p3p3(p2+q2+pq)=q(p2−p+1)p3(1+3pq)=q(p2−p+1)p3(3p2+3p+1)=1+p33p5+3p4=1p4(p+1)=13⇒3(t+1)4t=1nowuseanypolynomialcalculator,say,wolframalpha.

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