If $\alpha ,\beta$  and $\gamma$  are the  roots of equation $x^3-3x^2+x+5=0$  then $y={\displaystyle \sum {\alpha }^2+\alpha \beta \gamma }$  satisfies the equation is

The given equation is $x^3-3x^2+x+5=0$

$\alpha ,\beta$  and $\gamma$  are the  roots of the equation

Then $\alpha +\beta +\gamma =3,\alpha \beta +\beta \gamma +\gamma \alpha =1,\alpha \beta \gamma =-5$

$y={\displaystyle \sum {\alpha }^2+\alpha \beta \gamma }$

$={\alpha }^2+{\beta }^2+{\gamma }^2+\alpha \beta \gamma$

Add & subtract $2\left(\alpha \beta +\beta \gamma +\gamma \alpha \right)$ , we get

$={\left(\alpha +\beta +\gamma \right)}^2-2\left(\alpha \beta +\beta \gamma +\gamma \alpha \right)+\alpha \beta \gamma$

$=9-2-5=2$

$\therefore$ $y=2$

Check $y=2$ in given options

It satisfies the equation $y^3-y^3-y-2=0$ .