Cubic root of unity

Question

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)

Answer 1

There is a 'trick' method: consider

$$6-\dfrac{2+17\omega}{2+3\omega} = \dfrac{12 + 18\omega-2-17\omega}{2+3\omega}=\dfrac{10+w}{2+3\omega}$$

Now if we multiply through by $\dfrac{\omega^2}{\omega^2}\dots$

Answer 2

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 $$ \frac{1+ 10w^2}{1-2w} + \frac{2+17w}{2+3w} = \frac{48+30w}{8+5w} = 6. $$

Answer 3

Using $w^2=-w-1$ gives you $$\begin{align}\frac{1+10w^2}{1-2w}+\frac{2+17w}{2+3w}&=\frac{(1+10(-w-1))(2+3w)+(2+17w)(1-2w)}{(1-2w)(2+3w)}\\&=\frac{-64w^2-34w-16}{2-w-6w^2}\\&=\frac{-64(-w-1)-34w-16}{2-w-6(-w-1)}\\&=\frac{6(5w+8)}{5w+8}\\&=6\end{align}$$