A seemingly polynomial equation

#Polynomial#Trigonometric Substitution

Problem Statement

Problem Content

$$ y+ x^3 = 3x(1+xy), $$

$$ z+y^3 = 3y(1+yz), $$

$$ x+z^3 = 3z(1+zx) $$

is equal to:

Upon rearranging these equations, we get the following.

$$ y = \frac{3x-x^3}{1-3x^2}, $$

$$ z = \frac{3y-y^3}{1-3y^2}, $$

$$ x = \frac{3z-z^3}{1-3z^2} $$

Since $x,y,z \in \mathbb{R},$ assume that $ x =\tan{\theta}. $

$$ \implies y = \frac{3 \tan{\theta}- \tan^3{\theta}}{1-3\tan^2{\theta}} = \tan{3 \theta} $$

$$ \text{Analogously}, \ z = \tan{9\theta}, \ \text{and} $$

$$ x = \tan{27 \theta} = \tan{\theta} $$

Solving the equation, we get $ 27 \theta = n \pi + \theta $

$$ 26 \theta = n \pi, \ \text{where} \ n \ \text{is an integer}. $$

$$ \theta = \frac{n \pi}{26}, \ n \ \in \lbrace 0,1,2, \dots 25 \rbrace. $$

In total, there are 26 solutions.