Radical Expression Simplification 2
The number $$\sqrt{104\sqrt6+468\sqrt{10}+144\sqrt{15}+2006}$$can be written as $ a\sqrt2+b\sqrt3+c\sqrt5 $, where $ a $, $ b $, and $ c $ are positive integers. Find $ a\cdot b\cdot c $.
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Solution
We have \[\left(a\sqrt{2}+b\sqrt{3}+c\sqrt{5}\right)^2 = 104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006,\]or \[2ab\sqrt{6}+2ac\sqrt{10}+2bc\sqrt{15}+(2a^2+3b^2+5c^2)=104\sqrt{6}+468\sqrt{10}+144\sqrt{15}+2006.\]Since $ a,b,c $ are integers, we get the four equations \[\begin{aligned} 2ab &= 104, \\ 2ac &= 468, \\ 2bc &= 144, \\ 2a^2+3b^2+5c^2 &= 2006.\end{aligned}\]Thus, $ ab=52 $, $ ac=234 $, and $ bc=72 $. To find $ abc $, we multiply these three equations together, getting \[(abc)^2 = 52 \cdot 234 \cdot 72 = (2^2 \cdot 13) \cdot (2 \cdot 3^2 \cdot 13) \cdot (2^3 \cdot 3^2) = 2^6 \cdot 3^4 \cdot 13^2.\]Then, $ abc = 2^3 \cdot 3^2 \cdot 13 = \boxed{936} $.
We can solve the equations $ ab = 52, $ $ ac = 234, $ and $ bc = 72 $ to get $ a = 13, $ $ b = 4, $ and $ c = 18, $ which do satisfy $ 2a^2 + 3b^2 + 5c^2 = 2006 $. Thus, such positive integers $ a, $ $ b, $ $ c $ do exist.
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