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

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(C) Given the cubic equation $x^3+px^2+qx+r=0$,the roots are $\alpha, \beta, \gamma$. From Vieta's formulas: $\alpha+\beta+\gamma = -p$ $\alpha\beta+\

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$3pq-3r-p^3$

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(C) Given the cubic equation $x^3+px^2+qx+r=0$,the roots are $\alpha, \beta, \gamma$. From Vieta's formulas: $\alpha+\beta+\gamma = -p$ $\alpha\beta+\beta\gamma+\gamma\alpha = q$ $\alpha\beta\gamma = -r$ We use the identity: $\alpha^3+\beta^3+\gamma^3-3\alpha\beta\gamma = (\alpha+\beta+\gamma)(\alpha^2+\beta^2+\gamma^2-\alpha\beta-\beta\gamma-\gamma\alpha)$ We know that $\alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha) = (-p)^2 - 2q = p^2-2q$. Substituting these into the identity: $\alpha^3+\beta^3+\gamma^3 - 3(-r) = (-p)((p^2-2q) - q)$ $\alpha^3+\beta^3+\gamma^3 + 3r = -p(p^2-3q)$ $\alpha^3+\beta^3+\gamma^3 = -p^3+3pq-3r$ $\alpha^3+\beta^3+\gamma^3 = 3pq-3r-p^3$.

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3+px^2+qx+r=0$,then $\alpha^3+\beta^3+\gamma^3=$

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