Question

lf $\alpha ,\beta ,\gamma$ are the roots of the equation $x^3-7x+7=0$, then the value of ${\alpha }^{-4}+{\beta }^{-4}+{\gamma }^{-4}$ is

• A. $\frac{11}{7}$
• B. $-\frac{11}{7}$
• C. $\frac{3}{7}$
• D. $-\frac{3}{7}$

Answer 1

Answer: (C)

$\frac{3}{7}$

Given equation is $x^3-7x+7=0$

$\alpha +\beta +\gamma =0$
$\alpha \beta +\beta \gamma +\gamma \alpha =-7$
$\alpha \beta \gamma =-7$

$W=\frac{1}{{\alpha }^4}+\frac{1}{{\beta }^4}+\frac{1}{{\gamma }^4}=\frac{(\alpha \beta {)}^4+({\beta \alpha )}^4+({\alpha \gamma )}^4}{{\left(\alpha \beta \gamma \right)}^4}$

$S=(\alpha \beta {)}^2+({\beta \alpha )}^2+(\gamma \beta {)}^2={(\alpha \beta +\beta \gamma +\gamma \alpha )}^2-2(\alpha \beta \gamma \left)\right(\alpha +\gamma +\beta )=49-0=49$
$D=({a+b+c)}^2-2(ab+bc+ca)=0-2.(-7)=14$

$W=S^2-2\left(49\right)\left(14\right)=49\times 21=1029$

$W=\frac{S^2-2\times 49\times 14}{7^4}=\frac{{49}^2-2.\times 49\times 14}{7^4}=\frac{1029}{7^4}=\frac{3}{7}$