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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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    @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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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, 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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    @AdamAndersson: Yeah, I could simplify it somewhat.2012-06-20