<s>[INST] Cubic root of unity Is there anyway to solve this without substituting with the values?
Prove that: $$\frac{1+10w^2}{1-2w} + \frac{2+17w}{2+3w} = 6$$. (Where $w$ & $w^2$ are the cubic roots of unity)
 [/INST] Using $w^2=-w-1$ gives you
$$\begin{align}\frac{1+10w^2}{1-2w}+\frac{2+17w}{2+3w}&=\frac{(1+10(-w-1))(2+3w)+(2+17w)(1-2w)}{(1-2w)(2+3w)}\\&=\frac{-64w^2-34w-16}{2-w-6w^2}\\&=\frac{-64(-w-1)-34w-16}{2-w-6(-w-1)}\\&=\frac{6(5w+8)}{5w+8}\\&=6\end{align}$$
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