MHB Is $\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}}$ a rational number?

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    2015
Euge
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Here's this week's problem!

________

Problem. Prove that $\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}}$ is rational.
________Note. The cube roots involved are principal cube roots.
Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!
 
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No one answered this week's problem correctly. You can find my solution below.

The sum is not only rational, but in fact, it equals $1$. Let $A = \sqrt[3]{2 + \sqrt{5}}$ and $B = \sqrt[3]{2 - \sqrt{5}}$. If $t = A + B$, then

$$t^3 = A^3 + B^3 + 3AB(A + B) = (2 + \sqrt{5}) + (2 - \sqrt{5}) + 3\sqrt[3]{-1}t = 4 - 3t.$$

Thus, $t$ is a real root of the cubic polynomial $x^3 + 3x - 4$. This polynomial factors as $(x - 1)(x^2 + x + 4)$ and $x^2 + x + 4$ has no real root, so $x^3 + 3x - 4$ has unique real root $1$. Therefore, $t = 1$.
 
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