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  4. If α ,β ,γ are the roots of the equation x3-7x+7=0, then (1/α 4)+(1/β 4)+(1/γ 4) is

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Q. If $ \alpha ,\beta ,\gamma $ are the roots of the equation $ {{x}^{3}}-7x+7=0, $ then $ \frac{1}{{{\alpha }^{4}}}+\frac{1}{{{\beta }^{4}}}+\frac{1}{{{\gamma }^{4}}} $ is

J & K CETJ & K CET 2007

Solution:

Given $ \alpha ,\beta ,\gamma $
are the roots of the equation
$ {{x}^{3}}-7x+7=0 $ .
$ \therefore $ $ \Sigma \alpha =0,\,\,\Sigma \alpha \beta =-7,\,\,\alpha \beta \gamma =-7 $
Now, $ \frac{1}{{{\alpha }^{4}}}+\frac{1}{{{\beta }^{4}}}+\frac{1}{{{\gamma }^{4}}}=\frac{{{\alpha }^{4}}{{\beta }^{4}}+{{\beta }^{4}}{{\gamma }^{4}}+{{\gamma }^{4}}{{\alpha }^{4}}}{{{\alpha }^{4}}{{\beta }^{4}}{{\gamma }^{4}}} $
$ =\frac{\Sigma {{\alpha }^{4}}{{\beta }^{4}}}{{{\alpha }^{4}}{{\beta }^{4}}{{\gamma }^{4}}} $ ..(i)
$ \Sigma \alpha \beta \,\Sigma \alpha \beta \,\Sigma \alpha \beta \,\Sigma \alpha \beta ={{(\Sigma \alpha \beta )}^{2}}.{{(\Sigma \alpha \beta )}^{2}} $
$ \Rightarrow $ $ {{(-7)}^{4}}=({{\alpha }^{2}}{{\beta }^{2}}+{{\beta }^{2}}{{\gamma }^{2}}+{{\gamma }^{2}}{{\alpha }^{2}}+2{{\alpha }^{2}}\beta \gamma $
$ +2\alpha {{\beta }^{2}}\gamma +2\alpha \beta {{\gamma }^{2}}) $
$ ({{\alpha }^{2}}{{\beta }^{2}}+{{\beta }^{2}}{{\gamma }^{2}}+{{\gamma }^{2}}{{\alpha }^{2}}+2{{\alpha }^{2}}{{\beta }^{2}}+2\alpha {{\beta }^{2}}\gamma +2\alpha \beta {{\gamma }^{2}}) $
$ =[{{\alpha }^{2}}{{\beta }^{2}}+{{\beta }^{2}}{{\gamma }^{2}}+{{\gamma }^{2}}{{\alpha }^{2}}+2\alpha \beta \gamma (\alpha +\beta +\gamma )] $
$ [{{\alpha }^{2}}{{\beta }^{2}}+{{\beta }^{2}}{{\gamma }^{2}}+{{\gamma }^{2}}{{\alpha }^{2}}+2\alpha \beta \gamma (\alpha +\beta +\gamma )] $
$ =({{\alpha }^{2}}{{\beta }^{2}}+{{\beta }^{2}}{{\gamma }^{2}}+{{\gamma }^{2}}{{\alpha }^{2}}) $
$ ({{\alpha }^{2}}{{\beta }^{2}}+{{\beta }^{2}}{{\gamma }^{2}}+{{\gamma }^{2}}{{\alpha }^{2}}) $
$ (\because \,\,\Sigma \alpha =\alpha +\beta +\gamma =0) $
$ ={{\alpha }^{2}}{{\beta }^{2}}+{{\beta }^{2}}{{\gamma }^{4}}+{{\gamma }^{4}}{{\alpha }^{4}}+2{{\alpha }^{4}}{{\beta }^{2}}{{\gamma }^{2}} $
$ +2{{\alpha }^{2}}{{\beta }^{4}}{{\gamma }^{2}}+2{{\alpha }^{2}}{{\beta }^{2}}{{\gamma }^{4}} $
$ =\Sigma {{\alpha }^{2}}{{\beta }^{4}}+2{{\alpha }^{2}}{{\beta }^{2}}{{\gamma }^{2}}({{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}}) $
$ =\Sigma {{\alpha }^{4}}{{\beta }^{4}}+2{{\alpha }^{2}}{{\beta }^{2}}{{\gamma }^{2}}[{{(\Sigma \alpha )}^{2}}-2\Sigma \alpha \beta ] $
$ =\Sigma {{\alpha }^{4}}{{\beta }^{4}}+2{{\alpha }^{2}}{{\beta }^{2}}{{\gamma }^{2}}[0-2\times (-7)] $
$ \Rightarrow $ $ {{(-7)}^{4}}=\Sigma {{\alpha }^{4}}{{\beta }^{4}}+2{{(-7)}^{2}}(2\times 7) $
$ \Rightarrow $ $ \Sigma {{\alpha }^{4}}{{\beta }^{4}}={{(-7)}^{4}}+4{{(-7)}^{3}} $
$ \Rightarrow $ $ \Sigma {{\alpha }^{2}}{{\beta }^{4}}={{(-7)}^{3}}(-7+4)=-3{{(-7)}^{3}} $
On putting this value in Eq. (i), we get $ \frac{1}{{{\alpha }^{4}}}+\frac{1}{{{\beta }^{4}}}+\frac{1}{{{\gamma }^{4}}}=\frac{-3{{(-7)}^{3}}}{{{(-7)}^{4}}}=\frac{-3}{-7}=\frac{3}{7} $ `