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Question Number 202374 by Calculusboy last updated on 25/Dec/23

Commented by Frix last updated on 25/Dec/23

x^3 +px+q=0 Σ_(k=1) ^3 (((1+x_k )^2 )/((1−x_k )^2 )) =((3p^2 +2pq+3q^2 −10p−30q+3)/(p+q+1)) Σ_(k=1) ^3 (((1−x_k )^2 )/((1+x_k )^2 )) =((3p^2 −2pq+3q^2 −10p+30q+3)/(p−q+1))

x3+px+q=0∑3k=1(1+xk)2(1−xk)2=3p2+2pq+3q2−10p−30q+3p+q+1∑3k=1(1−xk)2(1+xk)2=3p2−2pq+3q2−10p+30q+3p−q+1

Commented by Calculusboy last updated on 26/Dec/23

nice solution

nicesolution

Answered by aleks041103 last updated on 25/Dec/23

x^3 −x−1 x=y+1 ⇒x^3 −x−1=(1+y)^3 −y−2= =y^3 +3y^2 +3y+1−y−2= =y^3 +3y^2 +2y−1=(y−(α−1))(y−(β−1))(y−(γ−1))= =(y−a)(y−b)(y−c) ⇒ { ((abc=1)),((ab+bc+ac=2)),((a+b+c=−3)) :} (((1+α)/(1−α)))^2 =(−(((1−α)−2)/(1−α)))^2 =(1+(2/(α−1)))^2 = =1+(4/(α−1))+(4/((α−1)^2 ))=1+(4/a)+(4/a^2 ) Σ(((1+α)/(1−α)))^2 =3+4Σ(1/a)+4Σ(1/a^2 ) Σ(1/a)=((ab+bc+ac)/(abc))=(2/1)=2 Σ(1/a^2 )=((a^2 b^2 +b^2 c^2 +a^2 c^2 )/((abc)^2 ))=(ab)^2 +(bc)^2 +(ac)^2 = =(ab+bc+ac)^2 −2(a^2 bc+ab^2 c+abc^2 )= =(ab+bc+ac)^2 −2abc(a+b+c)= =2^2 −2(−3)=10 ⇒Σ(((1+α)/(1−α)))^2 =3+4.2+4.10=51 ⇒Σ_(α,β,γ) (((1+x)/(1−x)))^2 =51, x^3 −x−1=0, ∀x∈{α,β,γ}

x3−x−1x=y+1⇒x3−x−1=(1+y)3−y−2==y3+3y2+3y+1−y−2==y3+3y2+2y−1=(y−(α−1))(y−(β−1))(y−(γ−1))==(y−a)(y−b)(y−c)⇒{abc=1ab+bc+ac=2a+b+c=−3(1+α1−α)2=(−(1−α)−21−α)2=(1+2α−1)2==1+4α−1+4(α−1)2=1+4a+4a2Σ(1+α1−α)2=3+4Σ1a+4Σ1a2Σ1a=ab+bc+acabc=21=2Σ1a2=a2b2+b2c2+a2c2(abc)2=(ab)2+(bc)2+(ac)2==(ab+bc+ac)2−2(a2bc+ab2c+abc2)==(ab+bc+ac)2−2abc(a+b+c)==22−2(−3)=10⇒Σ(1+α1−α)2=3+4.2+4.10=51⇒∑α,β,γ(1+x1−x)2=51,x3−x−1=0,∀x∈{α,β,γ}

Commented by Calculusboy last updated on 25/Dec/23

thanks sir

thankssir

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