Question

If $\alpha$ and $\beta$ are the roots of the equation $x^2+2x+4=0$, then $\frac{1}{{\alpha }^3}+\frac{1}{{\beta }^3}$ is equal to

• A. $-1/2$
• B. $1/2$
• C. $32$
• D. $1/4$

Answer 1

Answer: (D)

Given equation is $x^2+2x+4=0$
Since $\alpha ,\beta$ are roots of this equation.
$\therefore \alpha +\beta =-2$ and $\alpha \beta =4$
Now, $\frac{1}{{\alpha }^3}+\frac{1}{{\beta }^3}$
$=\frac{{\alpha }^3+{\beta }^3}{{\left(\alpha \beta \right)}^3}$
$=\frac{\left(\alpha +\beta \right)\left({\alpha }^2+{\beta }^2-\alpha \beta \right)}{{\left(\alpha \beta \right)}^3}$
$=\frac{\left(-2\right)\left({\left(\alpha +\beta \right)}^2-3\alpha \beta \right)}{4\times 4\times 4}$
$=\frac{-2\left(4-12\right)}{4\times 4\times 4}$
$=\frac{\left(-2\right)\times \left(-8\right)}{4\times 4\times 4}=\frac{1}{4}$