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Question Number 90709 by Cynosure last updated on 25/Apr/20

α,β and γ are the roots of x^3 −9x+9=0 find the value of (1) α^(−3) +β^(−3) +γ^(−3) (2) α^(−5) +β^(−5) +γ^(−5)

$$\alpha,\beta\:{and}\:\gamma\:{are}\:{the}\:{roots}\:{of}\:\:{x}^{\mathrm{3}} −\mathrm{9}{x}+\mathrm{9}=\mathrm{0} \\ $$$${find}\:{the}\:{value}\:{of}\:\left(\mathrm{1}\right)\:\alpha^{−\mathrm{3}} +\beta^{−\mathrm{3}} +\gamma^{−\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{2}\right)\:\alpha^{−\mathrm{5}} +\beta^{−\mathrm{5}} +\gamma^{−\mathrm{5}} \\ $$

Commented by jagoll last updated on 25/Apr/20

let α^3 = t ⇒ α = (t)^(1/(3 )) ⇒ ((t)^(1/(3 )) )^3 +9 = 9 (t)^(1/(3 )) (t+9)^3 = 729t ⇒ t^3 +27t^2 +243t+729 = 729t t^3 +27t^2 −486t+729 = 0 , has roots α^3 , β^3 & γ^3 find the polynomial with roots (1/α^3 ) , (1/β^3 ) & (1/γ^3 ) ⇒ 729x^3 −486x^2 +27x+1 = 0 by vieta rule (1/α^3 ) + (1/β^3 )+(1/γ^3 ) = ((486)/(729)) = ((162)/(243)) =((54)/(81)) = (2/3)

$${let}\:\alpha^{\mathrm{3}} \:=\:{t}\:\Rightarrow\:\alpha\:=\:\sqrt[{\mathrm{3}\:\:}]{{t}} \\ $$$$\Rightarrow\:\left(\sqrt[{\mathrm{3}\:\:}]{{t}}\:\right)^{\mathrm{3}} +\mathrm{9}\:=\:\mathrm{9}\:\sqrt[{\mathrm{3}\:\:}]{{t}} \\ $$$$\left({t}+\mathrm{9}\right)^{\mathrm{3}} \:=\:\mathrm{729}{t}\:\Rightarrow\:{t}^{\mathrm{3}} +\mathrm{27}{t}^{\mathrm{2}} +\mathrm{243}{t}+\mathrm{729}\:=\:\mathrm{729}{t} \\ $$$${t}^{\mathrm{3}} +\mathrm{27}{t}^{\mathrm{2}} −\mathrm{486}{t}+\mathrm{729}\:=\:\mathrm{0}\:,\:{has}\:{roots} \\ $$$$\alpha^{\mathrm{3}} \:,\:\beta^{\mathrm{3}} \:\&\:\gamma^{\mathrm{3}} \\ $$$${find}\:{the}\:{polynomial}\:{with}\:{roots} \\ $$$$\frac{\mathrm{1}}{\alpha^{\mathrm{3}} }\:,\:\frac{\mathrm{1}}{\beta^{\mathrm{3}} }\:\&\:\frac{\mathrm{1}}{\gamma^{\mathrm{3}} }\:\Rightarrow\:\mathrm{729}{x}^{\mathrm{3}} −\mathrm{486}{x}^{\mathrm{2}} +\mathrm{27}{x}+\mathrm{1}\:=\:\mathrm{0} \\ $$$${by}\:{vieta}\:{rule}\:\frac{\mathrm{1}}{\alpha^{\mathrm{3}} }\:+\:\frac{\mathrm{1}}{\beta^{\mathrm{3}} }+\frac{\mathrm{1}}{\gamma^{\mathrm{3}} }\:=\:\frac{\mathrm{486}}{\mathrm{729}}\:=\:\frac{\mathrm{162}}{\mathrm{243}} \\ $$$$=\frac{\mathrm{54}}{\mathrm{81}}\:=\:\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$ \\ $$

Commented by Cynosure last updated on 25/Apr/20

what of the second one sir?

$${what}\:{of}\:{the}\:{second}\:{one}\:{sir}? \\ $$

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