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June 22nd, 2014, 01:24 AM   #1
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June 22nd, 2014, 05:53 AM   #2
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$\displaystyle \sqrt[\large3]{45+29\sqrt2} + \sqrt[\large3]{45-29\sqrt2} = (3 + \sqrt2) + (3 - \sqrt2) = 6$
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June 22nd, 2014, 06:05 AM   #3
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Let $\displaystyle A= \sqrt[3]{x} + \sqrt[3]{y}$ where $\displaystyle x=45+29\sqrt{2}$ and $\displaystyle y=45-29\sqrt{2}$.

Note that $\displaystyle x+y=90$, $\displaystyle xy=2025-841 \cdot 2=343=7^3$

Then

$\displaystyle A^3 = (\sqrt[3]{x} + \sqrt[3]{y})^3$
$\displaystyle A^3 = x + 3\sqrt[3]{x^2y}+3\sqrt[3]{xy^2}+y$
$\displaystyle A^3 = (x+y) + 3\sqrt[3]{xy}(\sqrt[3]{x} + \sqrt[3]{y})$
$\displaystyle A^3 = 90 + 21A$
$\displaystyle A^3 - 21A - 90=0$

This equation has two complex roots and one real real. Using the rational roots theorem we find that the only real root is 6.

Thus $\displaystyle \sqrt[3]{45+29\sqrt{2}} + \sqrt[3]{45-29\sqrt{2}}=6$ and is therefore a rational number.
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