Question

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

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

Answer 1

Answer: (B)

$\alpha +\beta +\gamma =0;\sum \alpha \beta =-2$ and $\alpha \beta \gamma =-3$
${\alpha }^3-2\alpha +3=0\Rightarrow {\alpha }^3=2\alpha -3$ .....(1)
${\Vert }^{1y}{\beta }^3=2\beta -3\text{ and }$ ....(2)
${\gamma }^3=2\gamma -3$ ....(3)
$\therefore {\alpha }^3+{\beta }^3=2(\alpha +\beta )-6$
${\alpha }^3+{\beta }^3+6=2(\alpha +\beta )$
$\therefore \frac{1}{{\alpha }^3+{\beta }^3+6}=\frac{1}{2(\alpha +\beta )}$
$\therefore \sum \frac{1}{{\alpha }^3+{\beta }^3+6}=\frac{1}{2}\left[\frac{1}{\alpha +\beta +\gamma -\gamma }+\frac{1}{\beta +\gamma +\alpha -\alpha }+\frac{1}{\gamma +\alpha +\beta -\beta }\right]$
$=\frac{-1}{2}\left[\frac{1}{\gamma }+\frac{1}{\alpha }+\frac{1}{\beta }\right](\alpha +\beta +\gamma =0)$
$=\frac{-1}{2}\left[\frac{\alpha \beta +\beta \gamma +\gamma \alpha }{\alpha \beta \gamma }\right]=\frac{-1}{2}\left[\frac{-2}{-3}\right]=\frac{-1}{3}$