Finding $\sqrt{(14+6\sqrt 5)^3}+\sqrt{(14-6\sqrt 5)^3}$

Question

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?

Answer 1

Note that $$14\pm 6\sqrt{5}=14\pm 2\sqrt{45}=9+5\pm 2\sqrt{9\times 5}=(\sqrt{9}\pm\sqrt{5})^2$$

Answer 2

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 $2^4\cdot 3^2 = \color{red}{144}.$