Question

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

• A. $\frac{7}{3}$
• B. $\frac{3}{7}$
• C. $\frac{4}{7}$
• D. $\frac{7}{4}$

Answer 1

Answer: (B)

$\frac{3}{7}$

Answer 2

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 E (i), we get $\frac{1}{{{\alpha }^{4}}}+\frac{1}{{{\beta }^{4}}}+\frac{1}{{{\gamma }^{4}}}=\frac{-3{{(-7)}^{3}}}{{{(-7)}^{4}}}=\frac{-3}{-7}=\frac{3}{7}$ `