2006 AIME I 第 12 题

Product-to-sum: $2\cos 4x \cos x = \cos 5x + \cos 3x,$ so the equation reads $y^3 + z^3 = (y + z)^3$

$(y+z)^3 - y^3 - z^3 = 3yz(y+z),$ so one of $\cos 3x,$ $\cos 5x,$ $\cos 4x \cos x$ must vanish

By the product-to-sum identity, $2 \cos 4x \cos x = \cos 5x + \cos 3x,$ so the right side is $(\cos 5x + \cos 3x)^3.$ Setting $y = \cos 3x$ and $z = \cos 5x,$ the equation becomes $y^3 + z^3 = (y + z)^3,$ and since $(y+z)^3 - y^3 - z^3 = 3yz(y + z),$ it holds exactly when $$\begin{aligned} \cos 3x &= 0, \\ \cos 5x &= 0, \end{aligned}$$ or $$\cos 4x \cos x = 0.$$

For $100 \lt x \lt 200$ in degrees: $\cos 3x = 0$ gives $x = 150;$ $\cos 5x = 0$ gives $x = 126,$ $162,$ $198;$ $\cos 4x = 0$ gives $x = 112.5,$ $157.5;$ and $\cos x = 0$ gives no solutions in the interval.

The sum is $$\begin{aligned} &150 + 126 + 162 \\ &\quad {}+ 198 + 112.5 + 157.5 \\ &= 906. \end{aligned}$$