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Q.
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
Complex Numbers and Quadratic Equations
Solution:
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, $\frac{1}{\alpha^{3}}+\frac{1}{\beta^{3}}$
$=\frac{\alpha^{3}+\beta^{3}}{\left(\alpha\beta\right)^{3}}$
$=\frac{\left(\alpha+\beta\right)\left(\alpha^{2}+\beta^{2}-\alpha\beta\right)}{\left(\alpha\beta\right)^{3}}$
$=\frac{\left(-2\right)\left(\left(\alpha+\beta\right)^{2}-3\alpha\beta\right)}{4\times4\times4}$
$=\frac{-2\left(4-12\right)}{4\times4\times4}$
$=\frac{\left(-2\right)\times\left(-8\right)}{4\times4\times4}=\frac{1}{4}$