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Let-a-be-the-unique-real-zero-of-x-3-x-1-find-the-simplest-possible-way-to-write-18-a-2-a-1-2-as-polynomial-expression-in-a-with-ratio-coefficients-




Question Number 210310 by Spillover last updated on 05/Aug/24
Let a be the unique real zero of x^3 +x+1.  find the simplest possible way to write   ((18)/((a^2 +a+1)^2 ))  as polynomial expression in  a  with ratio coefficients
Letabetheuniquerealzeroofx3+x+1.findthesimplestpossiblewaytowrite18(a2+a+1)2aspolynomialexpressioninawithratiocoefficients
Commented by Frix last updated on 07/Aug/24
Do you know the answer?
Doyouknowtheanswer?
Commented by AlagaIbile last updated on 07/Aug/24
  6(a^2  + 1)^3
6(a2+1)3
Commented by Frix last updated on 08/Aug/24
((18)/((a^2 +a+1)^2 ))≈29.34  6(a^2 +1)^3 ≈18.89
18(a2+a+1)229.346(a2+1)318.89
Commented by Spillover last updated on 08/Aug/24
14a^2 −10a+16
14a210a+16
Answered by Frix last updated on 08/Aug/24
We have a^3 +a+1=0∧a∈R  ((18)/((a^2 +a+1)^2 ))=((18)/(a^4 +2a^3 +3a^2 +2a+1))=       a^3 =−(a+1) ⇒ a^4 =−a(a+1)  =((18)/(2a^3 +2a^2 +a+1))=       a^3 =−(a+1)  =((18)/(2a^2 −a−1))=       a^2 =−((a+1)/a)  =−((18a)/(a^2 +3a+2))=−((18a)/((a+1)(a+2)))=  =((18)/(a+1))−((36)/(a+2))  We know a=u^(1/3) +v^(1/3)   ω=−(1/2)+((√3)/2)i  (p/(a+q))=(p/(q+u^(1/3) +v^(1/3) ))=  =(p/(q+u^(1/3) +v^(1/3) ))×(((q+ωu^(1/3) +ω^2 v^(1/3) )(q+ω^2 u^(1/3) +ωv^(1/3) ))/((q+ωu^(1/3) +ω^2 v^(1/3) )(q+ω^2 u^(1/3) +ωv^(1/3) )))=  =((p(q^2 −(u^(1/3) +v^(1/3) )q+u^(2/3) −u^(1/3) v^(1/3) +v^(2/3) ))/(q^3 +u+v−3u^(1/3) v^(1/3) q))=       Cardano gives u, v:       a=(−(1/2)+((√(93))/(18)))^(1/3) +(−(1/2)−((√(93))/(18)))^(1/3)        u=−(1/2)+((√(93))/(18))∧v=−(1/2)−((√(93))/(18))       uv=−(1/(27)) ⇒ u+v=−1∧u^(1/3) v^(1/3) =−(1/3)  =((p(q^2 −qa+u^(2/3) +2u^(1/3) v^(1/3) +v^(2/3) −3u^(1/3) v^(1/3) ))/(q^3 +q−1))=  =((p(q^2 −qa+a^2 +1))/(q^3 +q−1))    ((18)/(a+1))−((36)/(a+2))=18(x^2 −x+2)−4(x^2 −2x+5)=  =2(7x^2 −5x+8)
Wehavea3+a+1=0aR18(a2+a+1)2=18a4+2a3+3a2+2a+1=a3=(a+1)a4=a(a+1)=182a3+2a2+a+1=a3=(a+1)=182a2a1=a2=a+1a=18aa2+3a+2=18a(a+1)(a+2)==18a+136a+2Weknowa=u13+v13ω=12+32ipa+q=pq+u13+v13==pq+u13+v13×(q+ωu13+ω2v13)(q+ω2u13+ωv13)(q+ωu13+ω2v13)(q+ω2u13+ωv13)==p(q2(u13+v13)q+u23u13v13+v23)q3+u+v3u13v13q=Cardanogivesu,v:a=(12+9318)13+(129318)13u=12+9318v=129318uv=127u+v=1u13v13=13=p(q2qa+u23+2u13v13+v233u13v13)q3+q1==p(q2qa+a2+1)q3+q118a+136a+2=18(x2x+2)4(x22x+5)==2(7x25x+8)
Answered by Spillover last updated on 08/Aug/24

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