Question

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

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

Answer: (C)

$4$

As $\alpha ,\beta ,\gamma$ are roots of $x^3+4x+1=0$
Gives
$s_1=\alpha +\beta +\gamma =0s_2=\alpha \beta +\beta \gamma +\gamma \alpha =4s_3=\alpha \beta \gamma =-1$
Now, $\alpha +\beta +\gamma =0\Rightarrow \alpha +\beta =-\gamma \Rightarrow \beta +\gamma =-\alpha \Rightarrow \alpha +\gamma =-\beta$
Therefore, ${(\alpha +\beta )}^{-1}+{(\beta +\gamma )}^{-1}+{(\alpha +\gamma )}^{-1}={(-\gamma )}^{-1}+{(-\alpha )}^{-1}+{(-\beta )}^{-1}$

$=\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma }$

$=\frac{-(\alpha \beta +\beta \gamma +\gamma \alpha )}{\alpha \beta \gamma }$

$=4$