Menu Close

Question-216919

Tinku Tara 0 Comments
Pin




Question Number 216919 by Abdullahrussell last updated on 24/Feb/25
Answered by Wuji last updated on 24/Feb/25
α^3 −3α^2 +5α−17=0−−−−−1 β^3 −3β^2 +5β+11=0−−−−−2 (1)+(2) α^3 +β^3 −3α^2 −3β^2 +5α+5β−6=0 α^3 +β^3 −3(α^2 +β^2 )+5(α+β)−6=0 α^2 +β^2 =(α+β)^2 −2αβ let α+β=S , αβ=P α^3 +β^3 =S(S^2 −3P) S(S^2 −3P)−3(S^2 −2P)+5S−6=0 S^3 −3S^2 −3SP+5S+6P−6=0 S^3 −3S^2 +5S−6−3SP+6P=0 S^3 −3S^2 +5S−6+P(−3S+6)=0 (S−2)(S^2 −S+3)+P(−3S+6)=0 (S−2)(S^2 −S+3)=0 S=2 Thus P(−3S+6) vanishes since S=2 S=α+β=2
α3−3α2+5α−17=0−−−−−1β3−3β2+5β+11=0−−−−−2(1)+(2)α3+β3−3α2−3β2+5α+5β−6=0α3+β3−3(α2+β2)+5(α+β)−6=0α2+β2=(α+β)2−2αβletα+β=S,αβ=Pα3+β3=S(S2−3P)S(S2−3P)−3(S2−2P)+5S−6=0S3−3S2−3SP+5S+6P−6=0S3−3S2+5S−6−3SP+6P=0S3−3S2+5S−6+P(−3S+6)=0(S−2)(S2−S+3)+P(−3S+6)=0(S−2)(S2−S+3)=0S=2ThusP(−3S+6)vanishessinceS=2S=α+β=2
Answered by Frix last updated on 25/Feb/25
x=α−1∧y=β−1 x^3 +2x−14=0 y^3 +2y+14=0 (x^3 +y^3 )+2(x+y)=0 (x+y)(x^2 −xy+y^2 +2)=0 x+y=0 α+β=2
x=α−1∧y=β−1x3+2x−14=0y3+2y+14=0(x3+y3)+2(x+y)=0(x+y)(x2−xy+y2+2)=0x+y=0α+β=2
Commented by Abdullahrussell last updated on 25/Feb/25
Sir, great solutions. Thanks.
Sir,greatsolutions.Thanks.

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *