If alpha, beta, gamma are the roots of the eq | vedclass

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(C) The given equation is $x^3 + 27 = 0$,which can be written as $x^3 = -27$.The roots are $x = -3, -3\omega, -3\omega^2$,where $\omega$ is the comple

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$x^2 + x + 1 = 0$

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(C) The given equation is $x^3 + 27 = 0$,which can be written as $x^3 = -27$.The roots are $x = -3, -3\omega, -3\omega^2$,where $\omega$ is the complex cube root of unity.Let $\alpha = -3$,$\beta = -3\omega$,and $\gamma = -3\omega^2$.Then $\frac{\beta}{\alpha} = \frac{-3\omega}{-3} = \omega$ and $\frac{\gamma}{\alpha} = \frac{-3\omega^2}{-3} = \omega^2$.The roots of the required quadratic equation are $\omega^2$ and $(\omega^2)^2 = \omega^4 = \omega$.The sum of the roots is $\omega^2 + \omega = -1$.The product of the roots is $\omega^2 \cdot \omega = \omega^3 = 1$.The quadratic equation is $x^2 - (	ext{sum of roots})x + (	ext{product of roots}) = 0$.Substituting the values,we get $x^2 - (-1)x + 1 = 0$,which simplifies to $x^2 + x + 1 = 0$.

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3 + 27 = 0$,find the quadratic equation whose roots are $\left( \frac{\gamma}{\alpha} \right)^2$ and $\left( \frac{\beta}{\alpha} \right)^2$.

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