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?
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Start with $(\sqrt{(14+6\sqrt{5})^3}+ \sqrt{(14-6\sqrt{5})^3})^2$ and use $(a+b)(a-b) = a^2-b^2$. – Martín-Blas Pérez Pinilla Nov 27 '15 at 10:57
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Note that $$14\pm 6\sqrt{5}=14\pm 2\sqrt{45}=9+5\pm 2\sqrt{9\times 5}=(\sqrt{9}\pm\sqrt{5})^2$$
mathlove
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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 $2^4\cdot 3^2 = \color{red}{144}.$
Jack D'Aurizio
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