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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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2  
One solution is $x = y = z = -1$. –  J. J. Jun 20 '12 at 15:43
    
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 $z$. –  Mark Bennet Jun 20 '12 at 15:51
    
Could you please give me a more detailed explanation. –  Adam Jun 20 '12 at 16:19
    
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). –  Mark Bennet Jun 20 '12 at 16:29
    
@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" –  Américo Tavares Jun 20 '12 at 20:15
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1 Answer 1

The only real solution is $x = y = z = -1$.

Claim 1: $x,y,z \ge -1$.

Proof. Suppose that $x < -1$. Then $0 = 2y^3 + 2x^3 + 3x + 3 > 2y^3 + 2$, so that $y < -1$ also. Similarly it follows that $z < -1$. Hence if one of $x,y,z$ is smaller than $-1$, all of them are. But then if for example $x<z$, we have $$0 = 2x^3 + 2z^2 + 3z + 3 < 2z^3 + 2z^2 + 3z + 3 = (z+1)(2z^2 + 3) < 0,$$ and we see that necessarily $x=y=z$, which implies that $x=y=z=-1$, contradiction.

Claim 2: $x,y,z \le -1$.

Proof. Suppose that $x > -1$ is the largest of $x,y,z$. So $z \le x$ and $$0 = 2x^3 + 2z^2 + 3z + 3 \ge 2z^3 + 2z^2 + 3z + 3 = (z+1)(2z^2 + 3),$$ which implies that $z \le - 1$. By Claim 1. $z = -1$ and hence also $x = -1$ and $y = -1$.

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That's the only real solution. There are also complex solutions, e.g. $x=y=z= \pm i \sqrt{6}/2$. –  Robert Israel Jun 20 '12 at 16:56
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Granted. I don't know why, but I was implicitly assuming that he wanted real solutions. –  J. J. Jun 20 '12 at 16:57
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Yeah, I am looking for real solutions only. Your proof is a bit too complicated. I am sure there is a simple proof that (-1, -1, -1) is the only real solution. –  Adam Jun 20 '12 at 19:42
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@AdamAndersson: Yeah, I could simplify it somewhat. –  J. J. Jun 20 '12 at 20:06
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