<s>[INST] Finding $\sqrt{(14+6\sqrt 5)^3}+\sqrt{(14-6\sqrt 5)^3}$ Find $$\sqrt{(14+6\sqrt{5})^3}+ \sqrt{(14-6\sqrt{5})^3}$$
A.$72$ 
B.$144$ 
C.$64\sqrt{5}$ 
D.$32\sqrt{5}$
How to cancel out the square root?
 [/INST] The square of that sum equals:
$$(14+6\sqrt{5})^3+(14-6\sqrt{5})^3+2\sqrt{(14^2-36\cdot 5)^3}$$
that is:
$$ 2\cdot 14^3 +2\cdot 3\cdot 14\cdot 36\cdot 5+2\sqrt{16^3} = 2^8\cdot 3^4$$
so the original sum equals $2^4\cdot 3^2 = \color{red}{144}.$
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