I was reading this article, where it is claimed (when I reformulate equation $(6)$ p.11) that the real root of $$ 3y^2 - 2y^3 -u= 0 $$ where $u \in (0,1)$, is $$ y = \frac{1}{2} - \frac{1}{2} \left[ \cos \left( \frac{\arccos(1-2u)}{3} \right) - \sqrt{3} \sin \left( \frac{\arccos(1-2u)}{3} \right) \right] $$ but how is this result obtained? I have tried using software like Wolfram and MATLAB but they do not give me results similar to this one. I would appreciate any help or suggestions.
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2Edited, My question is more about how this result can be obtained, because in the article they only mention that the solution of the polynomial is that, but I am interested in knowing how the result is obtained. Regards – daniel Dec 31 '24 at 15:02
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1This question is similar to: Inverse function of y = -2x^3 + 3x^2 (see the end of Claude's answer to this older post), and some other posts. – Anne Bauval Dec 31 '24 at 15:23