# 2006 AIME I Problems/Problem 12

## Problem

Find the sum of the values of $x$ such that $\cos^3 3x+ \cos^3 5x = 8 \cos^3 4x \cos^3 x$, where $x$ is measured in degrees and $100< x< 200.$

## Solution

Observe that $2\cos 4x\cos x = \cos 5x + \cos 3x$ by the sum-to-product formulas. Defining $a = \cos 3x$ and $b = \cos 5x$, we have $a^3 + b^3 = (a+b)^3 \rightarrow ab(a+b) = 0$. But $a+b = 2\cos 4x\cos x$, so we require $\cos x = 0$, $\cos 3x = 0$, $\cos 4x = 0$, or $\cos 5x = 0$.

Hence we see by careful analysis of the cases that the solution set is $A = \{150, 126, 162, 198, 112.5, 157.5\}$ and thus $\sum_{x \in A} x = \boxed{906}$.

## See also

 2006 AIME I (Problems • Answer Key • Resources) Preceded byProblem 11 Followed byProblem 13 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions

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