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(C) Given that $\alpha, \beta, \gamma$ are the roots of $x^3+ax^2+bx+c=0$. By Vieta's formulas: $\alpha+\beta+\gamma = -a$ $\alpha\beta+\beta\gamma+\g

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$\alpha^2, \beta^2, \gamma^2$

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(C) Given that $\alpha, \beta, \gamma$ are the roots of $x^3+ax^2+bx+c=0$. By Vieta's formulas: $\alpha+\beta+\gamma = -a$ $\alpha\beta+\beta\gamma+\gamma\alpha = b$ $\alpha\beta\gamma = -c$ Let the roots of the new equation $x^3+(2b-a^2)x^2+(b^2-2ac)x-c^2=0$ be $\alpha', \beta', \gamma'$. Comparing coefficients: $\alpha'+\beta'+\gamma' = -(2b-a^2) = a^2-2b = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha) = \alpha^2+\beta^2+\gamma^2$. $\alpha'\beta'+\beta'\gamma'+\gamma'\alpha' = b^2-2ac = (\alpha\beta+\beta\gamma+\gamma\alpha)^2 - 2(\alpha+\beta+\gamma)(\alpha\beta\gamma) = \alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2$. $\alpha'\beta'\gamma' = c^2 = (\alpha\beta\gamma)^2 = \alpha^2\beta^2\gamma^2$. Thus,the roots are $\alpha^2, \beta^2, \gamma^2$.

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

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