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(D) Given the cubic equation $x^3+a x^2-b x+c=0$ with roots $\alpha, \beta, \gamma$. By Vieta's formulas: $\alpha+\beta+\gamma = -a$ $\alpha \beta+\be

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$a b+3 c$

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(D) Given the cubic equation $x^3+a x^2-b x+c=0$ with roots $\alpha, \beta, \gamma$. By Vieta's formulas: $\alpha+\beta+\gamma = -a$ $\alpha \beta+\beta \gamma+\gamma \alpha = -b$ $\alpha \beta \gamma = -c$ We need to evaluate $\sum \beta^2(\gamma+\alpha) = \beta^2(\gamma+\alpha) + \gamma^2(\alpha+\beta) + \alpha^2(\beta+\gamma)$. Since $\alpha+\beta+\gamma = -a$,we have $(\gamma+\alpha) = -a-\beta$,$(\alpha+\beta) = -a-\gamma$,and $(\beta+\gamma) = -a-\alpha$. Substituting these: $\sum \beta^2(-a-\beta) = -a(\alpha^2+\beta^2+\gamma^2) - (\alpha^3+\beta^3+\gamma^3)$. Alternatively,using the identity $\sum \alpha^2(\beta+\gamma) = (\alpha+\beta+\gamma)(\alpha \beta+\beta \gamma+\gamma \alpha) - 3 \alpha \beta \gamma$: $= (-a)(-b) - 3(-c) = ab + 3c$.

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3+a x^2-b x+c=0$,then $\sum \beta^2(\gamma+\alpha) = $

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