I am trying to solve this system of equations in the set of all real numbers. Solve the system of equations $$\begin{cases} x^3+y (y-z)^2=2, \\ y^3+z(z-x)^2=3, \\ z^3+x(x-y)^2=8. \end{cases}$$ I tried From the given system, we have $$x^3 + y (y - z)^2 ) + 2 (y^3 + z (z - x)^2) - z^3 - x (x - y)^2 = 0.$$ I am trying to factor it. But I can not. How to solve this system of equation?
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2What’s the context? Are you trying to find all solutions, or some solutions? Are the variables supposed to be integers, rational numbers, real numbers or complex ones? – Alon Amit Jan 17 '22 at 02:11
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There’s a very simple solution in integers which you can find by inspection, or trial and error. – Alon Amit Jan 17 '22 at 02:12
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I want to find all solutions in real numbers . – minhthien_2016 Jan 17 '22 at 02:36
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@minhthien_2016 Any other conditions? – Jan 17 '22 at 03:17
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@RamanujanXV There is no other conditions. I used Mathematica and get unique solution $(1, 1, 2)$. – minhthien_2016 Jan 17 '22 at 03:30
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You face a polynomial of degree $27$ with only one real root – Claude Leibovici Jan 17 '22 at 04:51