$\text{ If }\alpha ,\beta \text{ and }\gamma \text{ are roots of }4x^3+8x^2-x-2=0,\text{ then the value of }\frac{4(\alpha +1)(\beta +1)(\gamma +1)}{\alpha \beta \gamma }\text{ is }$

$\begin{array}{l}4x^3+8x^2-x-2=4(x-\alpha )(x-\beta )(x-\gamma )\\ \text{ Putting }x=-1\text{, we get }\\ 3=4(-1-\alpha )(-1-\beta )(-1-\gamma )\\ \text{ or }(\alpha +1)(\beta +1)(\gamma +1)=-\frac{3}{4}\end{array}$

$\text{ Also, }\alpha \beta \gamma =\frac{1}{2}$

$\therefore \frac{4(\alpha +1)(\beta +1)(\gamma +1)}{\alpha \beta \gamma }=-6$