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At the end of the following article:

http://www.ijpam.eu/contents/2013-85-1/15/15.pdf

It is asserted that the russian mathematician, Sergey Markelov, in private communication, told them that he discovered the following trigonometric identity

$\sqrt[3]{\sin \frac{2\pi}{7}} + \sqrt[3]{\sin \frac{4\pi}{7}} + \sqrt[3]{\sin \frac{8\pi}{7}} = \sqrt[3]{\frac{\sqrt[3]{7}}{3} - 2 +\sqrt[3]{3\sqrt[3]{7} - 4}+\sqrt[3]{3\sqrt[3]{7} - 5}}\sqrt[3]{\frac{3}{2}\sqrt[6]{7}}$

However, there isn't further reference about this remarkable identity. How to prove it?

  • 1
    Cubic Gauss sums, I would say. – Jack D'Aurizio Sep 21 '17 at 22:38
  • 1
    Can you tell me, please, if $\sqrt[3]{\frac{3}{2}\sqrt[6]{7}}$ is a second factor in RHS (in which case it would be better for the reader to put it as first factor) or it must be inside the large cubic root? – Piquito Sep 26 '17 at 21:04
  • @Fractional Inquirer: You do not have to answer me any more. I already found out that it is a factor. – Piquito Sep 26 '17 at 21:22
  • Tito Piezas III gave a general solution to these kinds of identities, and Ramanujan also gave a less general solution. But it only works for “simple” trigonometric identities, and cubics. – Crescendo Oct 12 '17 at 16:53
  • 1
    Related: https://math.stackexchange.com/q/644003/ – Grigory M Nov 10 '17 at 21:46
  • The link presented in the article is now obsolete.I tried to find Sergey Markelov's contact information, but alas, couldn't find anything. He appears to be a mathematics educator, not a researcher – Yuriy S Apr 10 '18 at 08:44

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