A) \[\frac{7}{3}\]
B) \[\frac{3}{7}\]
C) \[\frac{4}{7}\]
D) \[\frac{7}{4}\]
Correct Answer: B
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 \[\frac{1}{{{\alpha }^{4}}}+\frac{1}{{{\beta }^{4}}}+\frac{1}{{{\gamma }^{4}}}=\frac{-3{{(-7)}^{3}}}{{{(-7)}^{4}}}=\frac{-3}{-7}=\frac{3}{7}\]
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