Question

If $\alpha ,\beta ,\gamma$are the roots of the equation $x^3+4x+1=0$, then ${\left(\alpha +\beta \right)}^{-1}+{\left(\beta +\gamma \right)}^{-1}+{\left(\gamma +\alpha \right)}^{-1}$ is:

• A. $2$
• B. $4$
• C. $3$
• D. $5$

Answer 1

The correct option is B

$4$

Step 1: Given information

If $\alpha ,\beta ,\gamma$are the roots of the equation $x^3+4x+1=0$,

Now,

$\begin{array}{rcl}{s}_1& =& \alpha +\beta +\gamma =0\\ s_2& =& \alpha \beta +\beta \gamma +\delta \alpha =4\\ s_3& =& \alpha \beta \gamma =-1\end{array}$

Now, here

$\begin{array}{rcl}\alpha +\beta +\gamma & =& 0\\ & \Rightarrow & \alpha +\beta =-\gamma \\ & \Rightarrow & \beta +\gamma =-\alpha \\ & \Rightarrow & \alpha +\gamma =-\beta \end{array}$

Step 2: Conclusion.

$\begin{array}{rcl}\text{Therefore},{\left(\alpha +\beta \right)}^{-1}+{\left(\beta +\gamma \right)}^{-1}+{\left(\alpha +\gamma \right)}^{-1}& =& {\left(-\gamma \right)}^{-1}+{\left(-\alpha \right)}^{-1}+{\left(-\beta \right)}^{-1}\end{array}$

$=\begin{array}{rcl}-\left(\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma }\right)& =& -\frac{\left(\alpha \beta +\beta \gamma +\gamma \alpha \right)}{\alpha \beta \gamma }=\frac{-4}{-1}=4\end{array}$

Hence, the correct option is (B).