June 22nd, 2014, 01:24 AM  #1 
Member Joined: Jun 2014 From: Math Forum Posts: 67 Thanks: 4  rational 
June 22nd, 2014, 05:53 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,805 Thanks: 2150 
$\displaystyle \sqrt[\large3]{45+29\sqrt2} + \sqrt[\large3]{4529\sqrt2} = (3 + \sqrt2) + (3  \sqrt2) = 6$

June 22nd, 2014, 06:05 AM  #3 
Senior Member Joined: Feb 2010 Posts: 706 Thanks: 141 
Let $\displaystyle A= \sqrt[3]{x} + \sqrt[3]{y}$ where $\displaystyle x=45+29\sqrt{2}$ and $\displaystyle y=4529\sqrt{2}$. Note that $\displaystyle x+y=90$, $\displaystyle xy=2025841 \cdot 2=343=7^3$ 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 $\displaystyle \sqrt[3]{45+29\sqrt{2}} + \sqrt[3]{4529\sqrt{2}}=6$ and is therefore a rational number. 

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