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Question Number 195765 by SajaRashki last updated on 10/Aug/23
hello { ((x^3 +(1/x^3 ) = 18)),((x>1)) :} ⇒ x^5 −(1/x^5 ) = ?
hello{x3+1x3=18x>1⇒x5−1x5=?
Answered by mr W last updated on 10/Aug/23
x^3 +(1/x^3 )=(x+(1/x))^3 −3(x+(1/x)) let s=x+(1/x) ∈R s^3 −3s−18=0 (s−3)(s^2 +3s+6)=0 ⇒s=3 (x^3 +(1/x^3 ))(x^2 +(1/x^2 ))=x^5 +(1/x^5 )+x+(1/x) x^5 +(1/x^5 )=(x^3 +(1/x^3 ))(x^2 +(1/x^2 ))−(x+(1/x)) x^5 +(1/x^5 )=(x^3 +(1/x^3 ))[(x+(1/x))^2 −2]−(x+(1/x)) x^5 +(1/x^5 )=18(s^2 −2)−s ⇒x^5 +(1/x^5 )=18(3^2 −2)−3=123 (x^5 −(1/x^5 ))^2 =(x^5 +(1/x^5 ))^2 −4 ⇒x^5 −(1/x^5 )=(√(123^2 −4))=55(√5) ✓
x3+1x3=(x+1x)3−3(x+1x)lets=x+1x∈Rs3−3s−18=0(s−3)(s2+3s+6)=0⇒s=3(x3+1x3)(x2+1x2)=x5+1x5+x+1xx5+1x5=(x3+1x3)(x2+1x2)−(x+1x)x5+1x5=(x3+1x3)[(x+1x)2−2]−(x+1x)x5+1x5=18(s2−2)−s⇒x5+1x5=18(32−2)−3=123(x5−1x5)2=(x5+1x5)2−4⇒x5−1x5=1232−4=555✓
Commented by SajaRashki last updated on 10/Aug/23
Mr. W; you are a genius in mathmatics thank you for nice your solution
Mr.W;youareageniusinmathmaticsthankyouforniceyoursolution
Commented by mr W last updated on 10/Aug/23
i′m not. but thank you anyway!
i′mnot.butthankyouanyway!
Answered by ajfour last updated on 10/Aug/23
say x^3 −(1/x^3 )=(x−(1/x))(x^2 +(1/x^2 )+1) =(√(324−4))=8(√5) x^5 −(1/x^5 )=Q =Q=(x−(1/x))(x^4 +(1/x^4 )+x^2 +(1/x^2 )+1) let x^2 +(1/x^2 )=t (Q/(8(√5)))=((t^2 −2+t+1)/(t+1)) x^3 +(1/x^3 )=18=(x+(1/x))(x^2 +(1/x^2 )−1) 324=(t+2)(t−1)^2 ⇒ t=7 (Q/(8(√5)))=((49−2+7+1)/8) ⇒ Q=x^5 −(1/x^5 )= 55(√5)
sayx3−1x3=(x−1x)(x2+1x2+1)=324−4=85x5−1x5=Q=Q=(x−1x)(x4+1x4+x2+1x2+1)letx2+1x2=tQ85=t2−2+t+1t+1x3+1x3=18=(x+1x)(x2+1x2−1)324=(t+2)(t−1)2⇒t=7Q85=49−2+7+18⇒Q=x5−1x5=555
Commented by SajaRashki last updated on 11/Aug/23
thank alot dear mr ajfour
thankalotdearmrajfour

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