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

$$\alpha ,\beta and\gamma aretherootsof{x}^3-9x+9=0$$$$findthevalueof\left(1\right){\alpha }^{-3}+{\beta }^{-3}+{\gamma }^{-3}$$$$\left(2\right){\alpha }^{-5}+{\beta }^{-5}+{\gamma }^{-5}$$

Commented by jagoll last updated on 25/Apr/20

$$let{\alpha }^3=t\Rightarrow \alpha =\sqrt[3]{t}$$$$\Rightarrow {\left(\sqrt[3]{t}\right)}^3+9=9\sqrt[3]{t}$$$${(t+9)}^3=729t\Rightarrow t^3+27t^2+243t+729=729t$$$$t^3+27t^2-486t+729=0,hasroots$$$${\alpha }^3,{\beta }^3\mathrm{\&}{\gamma }^3$$$$findthepolynomialwithroots$$$$\frac{1}{{\alpha }^3},\frac{1}{{\beta }^3}\mathrm{\&}\frac{1}{{\gamma }^3}\Rightarrow 729x^3-486x^2+27x+1=0$$$$byvietarule\frac{1}{{\alpha }^3}+\frac{1}{{\beta }^3}+\frac{1}{{\gamma }^3}=\frac{486}{729}=\frac{162}{243}$$$$=\frac{54}{81}=\frac{2}{3}$$$$$$

Commented by Cynosure last updated on 25/Apr/20

$$whatofthesecondonesir?$$

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