<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] Note that 
$1+w+w^2 = 0$, as $$w(1+w+w^2) = w+w^2+w^3 = 1+w+w^2$$
 and $w \ne 1$. Now we have
$$ (1-2w)(2+3w) = 2 - w - 6w^2 = 2 - w - 6(-1-w) = 8+5w $$
and 
\begin{align*}
  (1+10w^2)(2+3w) + (1-2w)(2+17w) &= 2+ 3w+20w^2 + 30 + 2 +13 w - 34w^2\\
        &= 34 + 16w - 14w^2\\
        &= 34 + 16w + 14 + 14w\\
        &= 48 + 30w
\end{align*} 
And therefore
$$
  \frac{1+ 10w^2}{1-2w} + \frac{2+17w}{2+3w} = \frac{48+30w}{8+5w} = 6.
$$
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