Question

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

• A. $\frac{a}{b}$
• B. $-\frac{a}{b}$
• C. $\frac{a^2}{b^2}$
• D. $-\frac{a^2}{b^2}$

Answer 1

Answer: (A)

$\frac{a}{b}$

As $\alpha ,\beta ,\gamma$ are roots of $x^3+ax+b=0$, gives
$S_1=\alpha +\beta +\gamma =0S_2=\alpha \beta +\beta \gamma +\gamma \alpha =a{S}_3=\alpha \beta \gamma =-b$
Now, $\alpha +\beta +\gamma =0\Rightarrow \alpha +\beta =-\gamma \Rightarrow \alpha +\gamma =-\beta \Rightarrow \beta +\gamma =-\alpha$
Therefore, ${(\alpha +\beta )}^{-1}+{(\beta +\gamma )}^{-1}+{(\gamma +\alpha )}^{-1}$

$=\frac{1}{(\alpha +\beta )}+\frac{1}{(\beta +\gamma )}+\frac{1}{(\gamma +\alpha )}=\frac{1}{-\gamma }+\frac{1}{-\alpha }+\frac{1}{-\beta }=-\left(\frac{\alpha \beta +\beta \gamma +\gamma \alpha }{\alpha \beta \gamma }\right)=-\frac{S_2}{S_3}=-\left(\frac{a}{-b}\right)$

$=\frac{a}{b}$