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(B) Given that $\alpha, \beta, \gamma$ are the roots of $x^3+p_1 x^2+p_2 x+p_3=0$. By Vieta's formulas,$p_1 = -(\alpha+\beta+\gamma) = -S_1 = -10$. $p

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$\frac{1910}{3}$

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(B) Given that $\alpha, \beta, \gamma$ are the roots of $x^3+p_1 x^2+p_2 x+p_3=0$. By Vieta's formulas,$p_1 = -(\alpha+\beta+\gamma) = -S_1 = -10$. $p_2 = \alpha\beta+\beta\gamma+\gamma\alpha = \frac{(\alpha+\beta+\gamma)^2 - (\alpha^2+\beta^2+\gamma^2)}{2} = \frac{S_1^2 - S_2}{2} = \frac{100-38}{2} = 31$. Using 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 have: $S_3 - 3(-p_3) = S_1(S_2 - p_2)$. $-1840 + 3p_3 = 10(38 - 31)$. $-1840 + 3p_3 = 10(7) = 70$. $3p_3 = 1910$. $p_3 = \frac{1910}{3}$.

If $\alpha, \beta, \gamma$ are the roots of the cubic equation $x^3+p_1 x^2+p_2 x+p_3=0$. Let $S_r=\alpha^r+\beta^r+\gamma^r$. Given $S_1=10, S_2=38$ and $S_3=-1840$,then $p_3=$

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