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)

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 $\frac{1}{{\alpha }^4}+\frac{1}{{\beta }^4}+\frac{1}{{\gamma }^4}=\frac{-3{(-7)}^3}{{(-7)}^4}=\frac{-3}{-7}=\frac{3}{7}$ `