The given equation implies that \[\cos^{3}3x+ \cos^{3}5x =(2\cos 4x\cos x)^3,\]and from the product-to-sum formula, $ 2 \cos 4x \cos x = \cos 5x + \cos 3x, $ so \[\cos^{3}3x+ \cos^{3}5x = (\cos5x+\cos 3x)^3.\]Let $ a=\cos 3x $ and $ b=\cos 5x $. Then $ a^3+b^3=(a+b)^3 $. Expand and simplify to obtain \[3ab(a + b) = 0.\]Thus, $ a=0, $ $ b=0, $ or $ a+b=0 $; that is, $ \cos 3x=0, $ $ \cos 5x=0, $ or $ \cos5x+\cos3x=0 $. The solutions to $ \cos 3x = 0 $ are of the form $ x=30^\circ+60^\circ j $, where $ j $ is an integer The solutions to $ \cos 5x = 0 $ are of the form $ x=18^\circ+36^\circ k $, where $ k $ is an integer. The equation $ \cos 3x + \cos 5x = 0 $ is equivalent to \[\cos4x\cos x=0,\]so its solutions are of the form $ x=22{1\over2}^\circ +45^\circ m $ and $ x=90^\circ +180^\circ n $, where $ m $ and $ n $ are integers. The solutions in the interval $ 100^\circ<x<200^\circ $ are $ 150^\circ, $ $ 126^\circ, $ $ 162^\circ, $ $ 198^\circ, $ $ 112{1\over2}^\circ $, and $ 157{1\over2}^\circ $, and their sum is $ \boxed{906} $ (in degrees).