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(D) Given the quadratic equation $x^2 + 2x + 4 = 0$.For a quadratic equation $ax^2 + bx + c = 0$,the sum of roots $\alpha + \beta = -\frac{b}{a}$ and

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$\frac{1}{4}$

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(D) Given the quadratic equation $x^2 + 2x + 4 = 0$.For a quadratic equation $ax^2 + bx + c = 0$,the sum of roots $\alpha + \beta = -\frac{b}{a}$ and the product of roots $\alpha \beta = \frac{c}{a}$.Here,$\alpha + \beta = -\frac{2}{1} = -2$ and $\alpha \beta = \frac{4}{1} = 4$.We need to find the value of $\frac{1}{\alpha^3} + \frac{1}{\beta^3}$.$\frac{1}{\alpha^3} + \frac{1}{\beta^3} = \frac{\alpha^3 + \beta^3}{(\alpha \beta)^3}$.Using the identity $\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha \beta(\alpha + \beta)$:$\frac{\alpha^3 + \beta^3}{(\alpha \beta)^3} = \frac{(\alpha + \beta)^3 - 3\alpha \beta(\alpha + \beta)}{(\alpha \beta)^3}$.Substituting the values:$= \frac{(-2)^3 - 3(4)(-2)}{(4)^3} = \frac{-8 + 24}{64} = \frac{16}{64} = \frac{1}{4}$.

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

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