Question

If $\alpha ,\beta ,\gamma$ are the roots of the cubic $x^3-2x+3=0$, then value of $\frac{1}{{\alpha }^3+{\beta }^3+6}+\frac{1}{{\beta }^3+{\gamma }^3+6}+\frac{1}{{\gamma }^3+{\alpha }^3+6}$ is equal to

• A. $\frac{1}{3}$
• B. $\frac{-1}{3}$
• C. $\frac{1}{2}$
• D. $\frac{-1}{2}$

Answer 1

The correct option is B $\frac{-1}{3}$

Since, $\alpha ,\beta ,\gamma$ are the roots of the equation
${\alpha }^3-2\alpha +3=0$
${\beta }^3-2\beta +3=0$
${\gamma }^3-2\gamma +3=0$
$\Rightarrow {\alpha }^3+{\beta }^3=2(\alpha +\beta )-6$
$\Rightarrow {\beta }^3+{\gamma }^3=2(\beta +\gamma )-6$
$\Rightarrow {\gamma }^3+{\alpha }^3=2(\gamma +\alpha )-6$
$\frac{1}{{\alpha }^3+{\beta }^3+6}+\frac{1}{{\beta }^3+{\gamma }^3+6}+\frac{1}{{\gamma }^3+{\alpha }^3+6}=\frac{1}{2(\alpha +\beta )}+\frac{1}{2(\beta +\gamma )}+\frac{1}{2(\gamma +\alpha )}$
$=\frac{1}{2(-\gamma )}+\frac{1}{2(-\alpha )}+\frac{1}{2(-\beta )}=\frac{\alpha \beta +\beta \gamma +\alpha \gamma }{-2\alpha \beta \gamma }$
$=\frac{-2}{-2(-3)}=\frac{-1}{3}$