Question

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

Answer 1

$\alpha ,\beta$ and $\gamma$ are roots of equation

$x^3+7x+7=0$

In cubic equation,

$a{x}^3+b{x}^2+cx+d=0$

Sum of roots$=-\frac{b}{a}$

Product of roots$=-\frac{d}{a}$

And, sum of product of roots $\frac{c}{a}$

So, here $\alpha +\beta +\gamma =-\frac{b}{a}=0$

$\alpha +\beta +\gamma =0\to \left(i\right)$

$\alpha \beta \gamma =-\frac{d}{a}=-7\to \left(ii\right)\alpha \beta +\beta \gamma +\gamma \alpha =\frac{c}{a}=7\to \left(iii\right)$

$\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma }=\frac{\alpha \beta +\beta \gamma +\gamma \alpha }{\alpha \beta \gamma }$

$\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma }=\frac{7}{-7}=-1$

Squaring both sides

$\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}+\frac{1}{{\gamma }^2}+\frac{2}{\alpha \beta }+\frac{2}{\beta \gamma }+\frac{2}{\gamma \alpha }=1\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}+\frac{1}{{\gamma }^2}+\frac{2(\alpha +\beta +\gamma )}{\alpha \beta \gamma }=1\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}+\frac{1}{{\gamma }^2}=1$

Squaring both sides

$\frac{1}{{\alpha }^4}+\frac{1}{{\beta }^4}+\frac{1}{{\gamma }^4}+2\frac{\left({{\alpha }^2{+\beta }^2+\gamma }^2\right)}{{\alpha }^2{\beta }^2{\gamma }^2}=1\frac{1}{{\alpha }^4}+\frac{1}{{\beta }^4}+\frac{1}{{\gamma }^4}=1-\frac{2[{\left({{\alpha }^2{+\beta }^2+\gamma }^2\right)}^2-2(\alpha \beta +\beta \gamma +\gamma \alpha )]}{{{\alpha }^2{+\beta }^2+\gamma }^2}\frac{1}{{\alpha }^4}+\frac{1}{{\beta }^4}+\frac{1}{{\gamma }^4}=1-\frac{2\left[2\left(7\right)\right]}{{(-7)}^2}=1+\frac{4}{7}\frac{1}{{\alpha }^4}+\frac{1}{{\beta }^4}+\frac{1}{{\gamma }^4}=\frac{11}{7}$