If $\alpha ,\beta ,\gamma$ are the roots of  $x^3+2x^2+3x+8=0,$

then$\left(\alpha +\beta \right)\left(\beta +\gamma \right)\left(\gamma +\alpha \right)=$

The given equation is $x^3+2x^2+3x+8=0$

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

Sum of the roots : $\alpha +\beta +\gamma =-2$

Sum of the product of two roots taken two at a time  : $\alpha \beta +\beta \gamma +\gamma \alpha =3$

Product of the roots : $\alpha \beta \gamma =-8$

Now$\left(\alpha +\beta \right)\left(\beta +\gamma \right)\left(\gamma +\alpha \right)=\left(\alpha +\beta +\gamma -\gamma \right)\left(\beta +\gamma +\alpha -\alpha \right)\left(\gamma +\alpha +\beta -\beta \right)$

                                                        $=\left(-2-\gamma \right)\left(-2-\alpha \right)\left(-2-\beta \right)$   $\left(\because \alpha +\beta +\gamma =-2\right)$

                                                    $=-\left[8+4\left(\alpha +\beta +\gamma \right)+2\left(\alpha \beta +\beta \gamma +\gamma \alpha \right)+\alpha \beta \gamma \right]$

                                                    $=-\left[8+4\left(-2\right)+2\left(3\right)+\left(-8\right)\right]$

                                                    $=-\left[-2\right]=2$

$\therefore \left(\alpha +\beta \right)\left(\beta +\gamma \right)\left(\gamma +\alpha \right)=2$