If $\alpha ,\beta$, and $\gamma$ are roots of $x^3-2x^2+6x-1=0$  the value of the expression 

$\alpha \left(\frac{{\alpha }^2+\alpha +1}{{\alpha }^2-\alpha +1}\right)+\beta \left(\frac{{\beta }^2+\beta +1}{{\beta }^2-\beta +1}\right)+\gamma \left(\frac{{\gamma }^2+\gamma +1}{{\gamma }^2-\gamma +1}\right)$ 

 $\frac{{\alpha }^2+\alpha +1}{{\alpha }^2-\alpha +1}=\frac{(\alpha -1)({\alpha }^2+\alpha +1)}{(\alpha -1)({\alpha }^2-\alpha +1)}=\frac{{\alpha }^3-1}{(\alpha -1)({\alpha }^2-\alpha +1)}$

$=\frac{2{\alpha }^2-6\alpha }{{\alpha }^3-2{\alpha }^2+2\alpha -1}=\frac{2\alpha (\alpha -3)}{-4\alpha }=\frac{3-\alpha }{2}$

Similarly remaining
$\frac{\alpha (3-\alpha )+\beta (3-\beta )+\gamma (3-\gamma )}{2}=\frac{3(\alpha +\beta +\gamma )-({\alpha }^2+{\beta }^2+{\gamma }^2)}{2}=7$