• # question_answer If a, b, g  are roots of equation ${{x}^{3}}+a{{x}^{2}}+bx+c=0$, then ${{\alpha }^{-1}}+{{\beta }^{-1}}+{{\gamma }^{-1}}=$ [EAMCET 2002] A) a/c B) - b/c C) b/a D) c/a

a, b, g are the roots of the equation      ${{x}^{3}}+a{{x}^{2}}+bx+c=0$ so $\alpha +\beta +\gamma =-a$, $\alpha \beta +\beta \gamma +\gamma \alpha =b$ and $\alpha \beta \gamma =-\,c$ Now${{\alpha }^{-1}}+{{\beta }^{-1}}+{{\gamma }^{-1}}$ $=\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma }$ $=\frac{\alpha \beta +\beta \gamma +\gamma \alpha }{\alpha \beta \gamma }$                                  $=-b/c$.