Proving that $\cot^220^\circ + \cot^240^\circ + \cot^280^\circ = 9$.

Question

Prove that $\cot^220^\circ + \cot^240^\circ + \cot^280^\circ = 9$.

I tried bringing them all to $\cot^220^\circ$ but it didn't work. How do I proceed?

Answer 1

$$\cot3x=\frac{\cot^3 x-3\cot x}{3\cot^2 x-1}$$ so $$\cot^23x=\frac{\cot^6 x-6\cot^4 x+9\cot^2x}{9\cot^4x-6\cot^2 x+1}.$$ Then for $a\in\{\pi/9,2\pi/9,4\pi/9\}$, $\cot^2a=1/3$. So these $\cot^2a$ are the roots of $$\frac{y^3-6y^2+9y}{9y^2-6y+1}=\frac13.$$ But this re-arranges to $y^3-9y^2+\cdots=0$. etc.