Question

If $\alpha ,\beta$ are the roots of $x^2-x+1=0$ then $\frac{{({\alpha }^2-\alpha )}^3+{({\beta }^2-\beta )}^3}{(2-\alpha )(2-\beta )}$ is equal to

• A. $0$
• B. $1$
• C. $\frac{-2}{3}$
• D. $2$

Answer 1

Answer: (C)

$\frac{-2}{3}$

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

${\alpha }^2-\alpha +1=0⟹{\alpha }^2-\alpha =-1$

${\beta }^2-\beta +1=0⟹{\beta }^2-\beta =-1$

$\alpha +\beta =1,\alpha \beta =1$

$(2-\alpha )(2-\beta )=4-2(\alpha +\beta )+\alpha \beta =4-2+1=3$

$\frac{({\alpha }^2-\alpha {)}^3+({\beta }^2-\beta {)}^3}{(2-\alpha )(2-\beta )}=\frac{(-1{)}^3+(-1{)}^3}{3}=-\frac{2}{3}$