Complex number (cube roots of unity) If $w$ and $w^2$ are non real cube roots of unity, then what would be the value of 
$\frac{2015+2016w+2017w^2}{2017+2015w+2016w^2}$+$\frac{2015+2016w+2017w^2}{2016+2017w+2015w^2}$ 
All I know is --> $1+w+w^2=0$ and $w^3=1$ 
Any tips and suggestions regarding how to solve these kinds of problems would be great. 

hint:$$\frac{2015+2016w+2017w^2}{2017+2015w+2016w^2}+\frac{2015+2016w+2017w^2}{2016+2017w+2015w^2}=\\
\frac{2015(1+w+w^2)+w+2w^2}{2015(w+1+w^2)+2+w^2}+\frac{2015(1+w+w^2)+w+2w^2}{2015(w^2+w+1)+1+2w}=\\
\frac{2015(0)+w+2w^2}{2015(0)+2+w^2}+\frac{2015(0)+w+2w^2}{2015(0)+1+2w}=\\
\frac{w+2w^2}{2+w^2}+\frac{w+2w^2}{1+2w}=\\
\frac{w(1+2w)}{2+w^2}+\frac{w(1+2w)}{1+2w}$$
so  :now put$w^2=-w-1$
$$\frac{w(1+2w)}{2+w^2}+\frac{w(1+2w)}{1+2w}=\\
\frac{w+w^2+w^2}{2+w^2}+w=\\
\frac{w+w^2+(-w-1)}{2+(-w-1)}+w=\\
\frac{w^2-1}{1-w}+w=\\
\frac{(w-1)(w+1)}{-(w-1)}+w=\\
-(w+1)+w=\\-1$$
