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(C) Let $f(x) = x^3 + \frac{a}{2}x + b = 0$. The roots are $\alpha, \beta, \gamma$. From Vieta's formulas,$\alpha+\beta+\gamma = 0$,$\alpha\beta+\beta

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$\frac{18 b^2}{a}$

Vedclass · Answer

(C) Let $f(x) = x^3 + \frac{a}{2}x + b = 0$. The roots are $\alpha, \beta, \gamma$. From Vieta's formulas,$\alpha+\beta+\gamma = 0$,$\alpha\beta+\beta\gamma+\gamma\alpha = \frac{a}{2}$,and $\alpha\beta\gamma = -b$. The roots of the second equation are $y_1 = (\alpha-\beta)(\alpha-\gamma)$,$y_2 = (\beta-\alpha)(\beta-\gamma)$,and $y_3 = (\gamma-\alpha)(\gamma-\beta)$. Note that $y_1 = \alpha^2 - \alpha(\beta+\gamma) + \beta\gamma = \alpha^2 - \alpha(-\alpha) + \beta\gamma = 2\alpha^2 + \beta\gamma$. Since $\alpha\beta\gamma = -b$,$\beta\gamma = -b/\alpha$. Thus $y_1 = 2\alpha^2 - b/\alpha = \frac{2\alpha^3-b}{\alpha}$. Since $\alpha^3 + \frac{a}{2}\alpha + b = 0$,we have $2\alpha^3 = -a\alpha - 2b$. Substituting this,$y_1 = \frac{-a\alpha - 2b - b}{\alpha} = -a - \frac{3b}{\alpha}$. Thus $y_1+a = -\frac{3b}{\alpha}$. Let $z = y+a$. Then $z_1 = -\frac{3b}{\alpha}, z_2 = -\frac{3b}{\beta}, z_3 = -\frac{3b}{\gamma}$. The equation for $z$ is $z^3 + Kz^2 + L = 0$. The roots are $z_1, z_2, z_3$. $K = -(z_1+z_2+z_3) = 3b(\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}) = 3b(\frac{\alpha\beta+\beta\gamma+\gamma\alpha}{\alpha\beta\gamma}) = 3b(\frac{a/2}{-b}) = -\frac{3a}{2}$. $L = -(z_1 z_2 z_3) = -(-\frac{27b^3}{\alpha\beta\gamma}) = -(\frac{27b^3}{b}) = -27b^2$. Wait,the equation is $z^3 + Kz^2 + L = 0$. The sum of roots taken two at a time is $0$. $z_1 z_2 + z_2 z_3 + z_3 z_1 = 9b^2(\frac{1}{\alpha\beta} + \frac{1}{\beta\gamma} + \frac{1}{\gamma\alpha}) = 9b^2(\frac{\alpha+\beta+\gamma}{\alpha\beta\gamma}) = 0$. So the equation is $z^3 + Kz^2 + L = 0$. $K = \frac{3a}{2}$ (adjusting signs). $L = -27b^2$. Re-evaluating: $L/K = \frac{-27b^2}{3a/2} = -18b^2/a$. Given the options,the magnitude is $18b^2/a$.

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3 + \frac{a}{2} x + b = 0$ and $(\alpha-\beta)(\alpha-\gamma)$,$(\beta-\alpha)(\beta-\gamma)$,$(\gamma-\alpha)(\gamma-\beta)$ are the roots of the equation $(y+a)^3 + K(y+a)^2 + L = 0$,then $\frac{L}{K} =$

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