Why $\sqrt[3]{{2 + \sqrt 5 }} + \sqrt[3]{{2 - \sqrt 5 }}$ is a rational number?
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1Related: https://math.stackexchange.com/questions/1416720, https://math.stackexchange.com/questions/1008169, https://math.stackexchange.com/questions/1180599/, https://math.stackexchange.com/questions/835955 – Watson Sep 25 '16 at 20:28
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Related: https://math.stackexchange.com/questions/2404139 – Watson Nov 22 '18 at 09:56
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Let $$\sqrt[3]{{2 + \sqrt 5 }} + \sqrt[3]{{2 - \sqrt 5 }}=x$$ $$(a+b)^3=a^3+b^3+3ab(a+b)$$ Then $$2 + \sqrt 5+2 - \sqrt 5-3x=x^3$$ $$x^3+3x=4$$ $$x=1$$ $$\sqrt[3]{{2 + \sqrt 5 }} + \sqrt[3]{{2 - \sqrt 5 }}=1$$
Roman83
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$3ab(a+b)=3\sqrt[3]{{2 + \sqrt 5 }} \cdot \sqrt[3]{{2 - \sqrt 5 }} (\sqrt[3]{{2 + \sqrt 5 }} + \sqrt[3]{{2 - \sqrt 5 }})=3(\sqrt[3]{4-5})x=-3x$ – Roman83 Sep 26 '16 at 05:26