Question

If $\alpha ,\beta$ the roots of $1+x+x^2=0$ then the value of ${\alpha }^4+{\beta }^4+{\alpha }^{-4}{\beta }^{-4}=$

Answer 1

Given,

$x^2+x+1=0$

$a=b=c=1$

$\alpha +\beta =-\frac{b}{a}=-1$

$\alpha \beta =\frac{c}{a}=1$

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

${\alpha }^4+{\beta }^4=({\alpha }^2+{\beta }^2{)}^2-2{\alpha }^2{\beta }^2=(-1{)}^2-2\left(1\right)=-1$

$\alpha \beta =1$

Given,

${\alpha }^4+{\beta }^4+{\alpha }^{-4}{\beta }^{-4}$

$=-1+\frac{1}{(1{)}^4}$

$=-1+1=0$