The roots of the equation x^3-3x^2+3x+7=0 are | vedclass

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(D) Given equation: $x^3-3x^2+3x+7=0$. This can be written as $(x-1)^3 + 8 = 0$,so $(x-1)^3 = -8$. Thus,$x-1 = -2, -2\omega, -2\omega^2$. The roots ar

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$3\omega^2$

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(D) Given equation: $x^3-3x^2+3x+7=0$. This can be written as $(x-1)^3 + 8 = 0$,so $(x-1)^3 = -8$. Thus,$x-1 = -2, -2\omega, -2\omega^2$. The roots are $\alpha = -1, \beta = 1-2\omega, \gamma = 1-2\omega^2$. Let $y = x-h$,so $x = y+h$. Substituting into the equation: $(y+h-1)^3 + 8 = 0$. For the $y^2$ and $y$ terms to be missing,we must have $h-1 = 0$,so $h=1$. The new roots are $\alpha-h = -2, \beta-h = -2\omega, \gamma-h = -2\omega^2$. We need to calculate $S = \frac{\alpha-h}{\beta-h} + \frac{\beta-h}{\gamma-h} + \frac{\gamma-h}{\alpha-h}$. $S = \frac{-2}{-2\omega} + \frac{-2\omega}{-2\omega^2} + \frac{-2\omega^2}{-2} = \frac{1}{\omega} + \frac{1}{\omega} + \omega^2 = \omega^2 + \omega^2 + \omega^2 = 3\omega^2$.

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