Question Number 174514 by Best1 last updated on 03/Aug/22

Commented by mr W last updated on 03/Aug/22

Answered by RedMath last updated on 03/Aug/22

Commented by Best1 last updated on 03/Aug/22

Answered by MJS_new last updated on 03/Aug/22

Commented by Tawa11 last updated on 03/Aug/22

Answered by behi834171 last updated on 03/Aug/22
![a^2 +ab+ac=a ab+b^2 +bc=b ac+bc+c^2 =c⇒2Σab=Σa−Σa^2 =1−2=−1 ⇒Σab=−(1/2) a^3 +a(b^2 +c^2 )=2a b(a^2 +c^2 )+b^3 =2b c(a^2 +b^2 )+c^3 =2c⇒ ⇒Σab(a+b)=2Σa−Σa^3 =2−3=−1 ⇒Σab(1−c)=−1⇒Σab−3abc=−1 ⇒−3abc=−1+(1/2)=−(1/2)⇒abc=(1/6) so:a,b,c, are the roots of equation: z^3 −z^2 −(1/2)z−(1/6)=0 z^5 =z^3 .z^2 =z^2 (z^2 +(1/2)z+(1/6))=z^4 +(1/2)z^3 +(1/6)z^2 = =z(z^2 +(1/2)z+(1/6))+(1/2)(z^2 +(1/2)z+(1/6))+(1/6)z^2 = =z^2 +(1/2)z+(1/6)+(1/2)z^2 +(1/6)z+(1/2)z^2 +(1/4)z+(1/(12))+(1/6)z^2 = =((13)/6)z^2 +((11)/(12))z+(1/4) ⇒Σa^5 =((13)/6)Σa^2 +((11)/(12))Σa+(3/4)= =((13)/6)×2+((11)/(12))×1+(3/4)=6 .■ ...................................... by using symbols of dear mr:MJS_new Σab=((Σa−Σa^2 )/2)=((𝛂−𝛃)/2) Σab−3abc=2Σa−Σa^3 ⇒ abc=−(1/3)(2𝛂−𝛄−((𝛂−𝛃)/2))=−((3𝛂+𝛃−2𝛄)/6) ⇒z^3 −𝛂z^2 +((𝛂−𝛃)/2)z+((3𝛂+𝛃−2𝛄)/6)=0 ⇒z^5 =z^2 z^3 =z^2 (𝛂z^2 −((𝛂−𝛃)/2)z−((3𝛂+𝛃−2𝛄)/6))= =𝛂z(𝛂z^2 −((𝛂−𝛃)/2)z−((3𝛂+𝛃−2𝛄)/6))−((𝛂−𝛃)/2)(𝛂z^2 −((𝛂−𝛃)/2)z−((3𝛂+𝛃−2𝛄)/6))−((3𝛂+𝛃−2𝛄)/6)z^2 = =𝛂^2 (𝛂z^2 −((𝛂−𝛃)/2)z−((3𝛂+𝛃−2𝛄)/6))−((𝛂(𝛂−𝛃))/2)z^2 −((𝛂(3𝛂+𝛃−2𝛄))/6)z− +((𝛂(𝛂−𝛃))/2)z^2 +(((𝛂−𝛃)^2 )/4)z+(((𝛂−𝛃)(3𝛂+𝛃−2𝛄))/(12))−((3𝛂+𝛃−2𝛄)/6)z^2 = =[𝛂^3 −((𝛂(𝛂−𝛃))/2)−((3𝛂+𝛃−2𝛄)/6)]z^2 −[((𝛂^2 (𝛂−𝛃))/2)+((𝛂(3𝛂+𝛃−2𝛄))/6)−(((𝛂−𝛃)^2 )/4)].z− −(((2𝛂^2 −𝛂+𝛃)(3𝛂+𝛃−2𝛄))/(12)) =A.z^2 +B.z+C A=((6𝛂^3 −3𝛂^2 +3𝛂𝛃−3𝛂−𝛃+2𝛄)/6) B=−((6𝛂^3 −6𝛂^2 𝛃+3𝛂^2 +8𝛂𝛃−4𝛂𝛄−3𝛃^2 )/(12)) C=−((6𝛂^3 +2𝛂^2 𝛃−4𝛂^2 𝛄−3𝛂^2 +2𝛂𝛃+2𝛂𝛄+𝛃^2 −2𝛃𝛄)/(12)) ⇒Σa^5 =A.Σa^2 +B.Σa+3C](https://www.tinkutara.com/question/Q174544.png)
Commented by Best1 last updated on 03/Aug/22

Commented by Tawa11 last updated on 03/Aug/22
