Question

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

• A. $0$
• B. $1$
• C. $-1$
• D. $3$

Answer 1

Answer: (A)

$0$

Given that,

$\alpha ,\beta$ are roots of the equation $x^2+x+1=0$

$\therefore \alpha +\beta =-1(sumofroots=\frac{(-1)co-efficientofx}{co-efficientof{x}^2})$

$\alpha \beta =1(productofroots=\frac{constantterm}{co-efficientof{x}^2})$

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

$\Rightarrow {\alpha }^2+{\beta }^2+2\alpha \beta =1$

$\Rightarrow {\alpha }^2+{\beta }^2+2=1$

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

Again squaring both sides,

$\Rightarrow {\alpha }^4+{\beta }^4+2{\alpha }^2{\beta }^2=1$

$\Rightarrow {\alpha }^4+{\beta }^4+2=1$

$\Rightarrow {\alpha }^4+{\beta }^4=-1$

$\therefore {\alpha }^4+{\beta }^4+\frac{1}{{\alpha }^4{\beta }^4}=-1+\frac{1}{1}=-1+1=0$