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

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    2015
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The discussion centers on proving that the expression $\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}}$ is a rational number. Participants are encouraged to explore the properties of cube roots and rationality. The problem remains unsolved by the community, prompting the original poster to share their solution. The focus is on the principal cube roots involved in the expression. The thread emphasizes the importance of following guidelines for problem-solving and sharing solutions.
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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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