If $$\alpha ,\beta ,\gamma $$  be the roots of the equation $$x\left( {1 + {x^2}} \right) + {x^2}\left( {6 + x} \right) + 2 = 0$$       then the value of $${\alpha ^{ - 1}} + {\beta ^{ - 1}} + {\gamma ^{ - 1}}$$    is

Solution :
$$2{x^3} + 6{x^2} + x + 2 = 0$$     has roots $$\alpha ,\beta ,\gamma .$$
So, $$2{x^3} + {x^2} + 6x + 2 = 0$$     has roots $${\alpha ^{ - 1}},{\beta ^{ - 1}},{\gamma ^{ - 1}}$$   (writing co-efficients in reverse order).
$$\therefore \,\,{\alpha ^{ - 1}} + {\beta ^{ - 1}} + {\gamma ^{ - 1}} = - \frac{1}{2}.$$