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(D) Given the cubic equation $2x^3+x^2-13x+6=0$. By Vieta's formulas,for roots $\alpha, \beta, \gamma$: $\alpha+\beta+\gamma = -\frac{1}{2}$ $\alpha\b

Vedclass · Accepted Answer

$-\frac{151}{8}$

Vedclass · Answer

(D) Given the cubic equation $2x^3+x^2-13x+6=0$. By Vieta's formulas,for roots $\alpha, \beta, \gamma$: $\alpha+\beta+\gamma = -\frac{1}{2}$ $\alpha\beta+\beta\gamma+\gamma\alpha = -\frac{13}{2}$ $\alpha\beta\gamma = -\frac{6}{2} = -3$ 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 $\alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha) = (-\frac{1}{2})^2 - 2(-\frac{13}{2}) = \frac{1}{4} + 13 = \frac{53}{4}$. Substituting these values: $\alpha^3+\beta^3+\gamma^3 = (\alpha+\beta+\gamma)(\frac{53}{4} - (-\frac{13}{2})) + 3(-3)$ $= (-\frac{1}{2})(\frac{53}{4} + \frac{26}{4}) - 9$ $= (-\frac{1}{2})(\frac{79}{4}) - 9$ $= -\frac{79}{8} - \frac{72}{8} = -\frac{151}{8}$.

If $\alpha, \beta, \gamma$ are the roots of the equation $2x^3+x^2-13x+6=0$,then $\alpha^3+\beta^3+\gamma^3=$

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