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I need help with the following system of equations:

$ 2y^3 +2x^2+3x+3=0 $

$ 2z^3 + 2y^2 + 3y + 3= 0 $

$2x^3 + 2z^2 + 3z + 3 = 0$

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    One solution is $x = y = z = -1$.2012-06-20
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    Since the equations are symmetric in $x, y, z$ it would be natural to try $x=y=z$ as a solution, which gives one real and two complex solutions. If you set $y=x$ you can work out what $z$ must be by adding and subtracting equations to extract a $(y-x)$ factor from all the terms in $x$ and $y$, leaving an equation in $z$.2012-06-20
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    Could you please give me a more detailed explanation.2012-06-20
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    I think that with homework you need to show that you are giving it a try. Put $x=y=z$ and see what you get. Then put $x=y$ and see what you get - actually what I did first isn't the easiest. Then you have the issue of whether the three variables can all be different. You might then want to think about which is largest (but you do not state whether you are looking for real solutions, where you can use the order, or complex ones where you can't).2012-06-20
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    @Adam Andersson Please incorporate in the question the information given in your comment: "I am looking for real solutions only. (...) "a simple proof that (-1, -1, -1) is the only real solution"2012-06-20

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