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

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(C) Given that $\alpha, \beta, \gamma$ are the roots of the equation $x^3+x^2+x+r=0$. From Vieta's formulas: $\alpha+\beta+\gamma = -1$ $\alpha\beta+\

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$-1$

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(C) Given that $\alpha, \beta, \gamma$ are the roots of the equation $x^3+x^2+x+r=0$. From Vieta's formulas: $\alpha+\beta+\gamma = -1$ $\alpha\beta+\beta\gamma+\gamma\alpha = 1$ $\alpha\beta\gamma = -r$ We use the identity: $\alpha^3+\beta^3+\gamma^3-3\alpha\beta\gamma = (\alpha+\beta+\gamma)(\alpha^2+\beta^2+\gamma^2-\alpha\beta-\beta\gamma-\gamma\alpha)$ Note that $\alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha) = (-1)^2 - 2(1) = 1-2 = -1$. Substituting the values into the identity: $5 - 3(-r) = (-1)(-1 - 1)$ $5 + 3r = (-1)(-2)$ $5 + 3r = 2$ $3r = 2 - 5$ $3r = -3$ $r = -1$

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3+x^2+x+r=0$ and $\alpha^3+\beta^3+\gamma^3=5$,then $r=$

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