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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

$$\alpha^{\mathrm{3}} −\mathrm{3}\alpha^{\mathrm{2}} +\mathrm{5}\alpha−\mathrm{17}=\mathrm{0}−−−−−\mathrm{1} \\ $$$$\beta^{\mathrm{3}} −\mathrm{3}\beta^{\mathrm{2}} +\mathrm{5}\beta+\mathrm{11}=\mathrm{0}−−−−−\mathrm{2} \\ $$$$\left(\mathrm{1}\right)+\left(\mathrm{2}\right) \\ $$$$\alpha^{\mathrm{3}} +\beta^{\mathrm{3}} −\mathrm{3}\alpha^{\mathrm{2}} −\mathrm{3}\beta^{\mathrm{2}} +\mathrm{5}\alpha+\mathrm{5}\beta−\mathrm{6}=\mathrm{0} \\ $$$$\alpha^{\mathrm{3}} +\beta^{\mathrm{3}} −\mathrm{3}\left(\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} \right)+\mathrm{5}\left(\alpha+\beta\right)−\mathrm{6}=\mathrm{0} \\ $$$$\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} =\left(\alpha+\beta\right)^{\mathrm{2}} −\mathrm{2}\alpha\beta \\ $$$$\mathrm{let}\:\alpha+\beta=\mathrm{S}\:,\:\alpha\beta=\mathrm{P} \\ $$$$\alpha^{\mathrm{3}} +\beta^{\mathrm{3}} =\mathrm{S}\left(\mathrm{S}^{\mathrm{2}} −\mathrm{3P}\right) \\ $$$$\mathrm{S}\left(\mathrm{S}^{\mathrm{2}} −\mathrm{3P}\right)−\mathrm{3}\left(\mathrm{S}^{\mathrm{2}} −\mathrm{2P}\right)+\mathrm{5S}−\mathrm{6}=\mathrm{0} \\ $$$$\mathrm{S}^{\mathrm{3}} −\mathrm{3S}^{\mathrm{2}} −\mathrm{3SP}+\mathrm{5S}+\mathrm{6P}−\mathrm{6}=\mathrm{0} \\ $$$$\mathrm{S}^{\mathrm{3}} −\mathrm{3S}^{\mathrm{2}} +\mathrm{5S}−\mathrm{6}−\mathrm{3SP}+\mathrm{6P}=\mathrm{0} \\ $$$$\mathrm{S}^{\mathrm{3}} −\mathrm{3S}^{\mathrm{2}} +\mathrm{5S}−\mathrm{6}+\mathrm{P}\left(−\mathrm{3S}+\mathrm{6}\right)=\mathrm{0} \\ $$$$\left(\mathrm{S}−\mathrm{2}\right)\left(\mathrm{S}^{\mathrm{2}} −\mathrm{S}+\mathrm{3}\right)+\mathrm{P}\left(−\mathrm{3S}+\mathrm{6}\right)=\mathrm{0} \\ $$$$\left(\mathrm{S}−\mathrm{2}\right)\left(\mathrm{S}^{\mathrm{2}} −\mathrm{S}+\mathrm{3}\right)=\mathrm{0} \\ $$$$\mathrm{S}=\mathrm{2} \\ $$$$\mathrm{Thus}\:\mathrm{P}\left(−\mathrm{3S}+\mathrm{6}\right)\:\mathrm{vanishes}\:\mathrm{since}\:\mathrm{S}=\mathrm{2} \\ $$$$\mathrm{S}=\alpha+\beta=\mathrm{2} \\ $$$$ \\ $$

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

$${x}=\alpha−\mathrm{1}\wedge{y}=\beta−\mathrm{1} \\ $$$${x}^{\mathrm{3}} +\mathrm{2}{x}−\mathrm{14}=\mathrm{0} \\ $$$${y}^{\mathrm{3}} +\mathrm{2}{y}+\mathrm{14}=\mathrm{0} \\ $$$$\left({x}^{\mathrm{3}} +{y}^{\mathrm{3}} \right)+\mathrm{2}\left({x}+{y}\right)=\mathrm{0} \\ $$$$\left({x}+{y}\right)\left({x}^{\mathrm{2}} −{xy}+{y}^{\mathrm{2}} +\mathrm{2}\right)=\mathrm{0} \\ $$$${x}+{y}=\mathrm{0} \\ $$$$\alpha+\beta=\mathrm{2} \\ $$

Commented by Abdullahrussell last updated on 25/Feb/25

Sir, great solutions. Thanks.

$$\:\:{Sir},\:{great}\:{solutions}.\:{Thanks}. \\ $$

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