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

$\begin{array}{l}\text{ Since }\alpha ,\beta ,\gamma \text{ are the roots of }{x}^3-2x+3=0\dots \dots \left(1\right)\\ \text{ Then }{S}_1=\alpha +\beta +\gamma =0,S_2=\alpha \beta +\beta \gamma +\gamma \alpha =-2,S_3=\alpha \beta \gamma =-3\end{array}$

$\text{ Substituting }\alpha ,\beta \text{ in equation (1), we get }{\alpha }^3=2\alpha -3,{\beta }^3=2\beta -3$

$\begin{array}{l}\therefore {\alpha }^3+{\beta }^3=2(\alpha +\beta )-6\\ \Rightarrow {\alpha }^3+{\beta }^3+6=2(\alpha +\beta )=2\left(-\gamma \right)\\ \therefore \sum \frac{1}{{\alpha }^3+{\beta }^3+6}=\sum -\frac{1}{2\gamma }\\ =-\frac{1}{2}\left[\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma }\right]\\ =-\frac{1}{2}\left[\frac{\alpha \beta +\beta \gamma +\gamma \alpha }{\alpha \beta \gamma }\right]\\ =-\frac{1}{2}\left[\frac{-2}{-3}\right]\\ =-\frac{1}{3}\end{array}$