If $\mathrm{\alpha },\mathrm{\beta },\mathrm{\gamma }$ are the roots of the cubic equation ${\mathrm{x}}^3+{\mathrm{P}}_1{\mathrm{x}}^2+{\mathrm{P}}_2{\mathrm{x}}^2+{\mathrm{P}}_3=0$ if 

${\mathrm{S}}_1={\mathrm{\alpha }}^{\mathrm{r}} +{\mathrm{\beta }}^{\mathrm{r}}+{\mathrm{\gamma }}^{\mathrm{r}}, {\mathrm{S}}_1=10, {\mathrm{S}}_2=38$ and ${\mathrm{S}}_3=-1840, \mathrm{then} {\mathrm{P}}_3=$

$\begin{array}{l}If\alpha ,\beta ,\gamma arerootsof{x}^3+p_1{x}^2+p_{2}x+p_3=0\\ \therefore x^3=-p_1{x}^2-p_{2}x-p_3--\left(1\right)\\ \therefore \alpha +\beta +\gamma =-p_1,\alpha \beta +\beta \gamma +\gamma \alpha =p_2,\alpha \beta \gamma =p_3\\ Given{s}_r={\alpha }^2+{\beta }^2+{\gamma }^2\\ s_1=10\Rightarrow \alpha +\beta +\gamma =10\\ \Rightarrow -p_1=10\Rightarrow p_1=-10\\ \therefore s_2=38\Rightarrow {\alpha }^2+{\beta }^2+{\gamma }^2=38\\ \Rightarrow {(\alpha +\beta +\gamma )}^2-2(\alpha \beta +\beta \gamma +\gamma \alpha )=38\\ \Rightarrow 100-2p_2=38\\ \Rightarrow 2p_2=100-38\\ \Rightarrow p_2=\frac{62}{2}\Rightarrow 31\\ from\left(1\right)\\ {\alpha }^3=-p_1{\alpha }^2-p_2\alpha -p_3\\ {\beta }^3=-p_1{\beta }^2-p_2\beta -p_3\\ {\gamma }^3=-p_1{\gamma }^2-p_2\gamma -p_3\\ \overline{{\alpha }^3+{\beta }^3+{\gamma }^3=-p_1({\alpha }^2+{\beta }^2+{\gamma }^2)-p_2(\alpha +\beta +\gamma )-3p_3}\\ \Rightarrow -1840=10\left(38\right)-31\left(10\right)-3p_3\\ -1840=70-3p_3\\ 3p_3=1910\\ \Rightarrow p_3=\frac{1910}{3}\end{array}$