How to solve this equation? [closed]

The numbers real x and y, are solutions of the following equation system :

$$\{x^3 + 3x^2y + 3xy^2 + y^3 2\sqrt2$$ $$\{x^3 - 3x^2 + 3xy^2 -2y^3=0$$ .

$$x^2 + y^2$$

How to solve it ?

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This does not appear to be a system of equations. What are the unmatched curly brackets meant to indicate? What's the significance of the first and last lines, given that they lack any relational operators? At a guess, you might try Solve[{x^3 + 3 x^2 y + 3 x y^2 + y^3 == 2 Sqrt[2], x^3 - 3 x^2 + 3 x y^2 - 2 y^3 == 0}, {x, y}] // First // Simplify. –  Oleksandr R. Sep 22 '12 at 5:27
I think downvotes are not necessary. This is Gustavo's first question in SE. Perhaps he needs some guiding and help to post a well formatted question ... but downvotes? Let's close this and ask him to post another meaningful one. –  belisarius Sep 22 '12 at 18:00
Gustavo, welcome to Mathematica.SE! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Read the FAQs! 3) When you see good Q&A, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. ALSO, remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign –  belisarius Sep 22 '12 at 20:39

closed as not a real question by whuber, Simon Woods, belisarius, Oleksandr R., Verbeia♦Sep 23 '12 at 1:44

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

Perhaps you are trying to deal with the following

ContourPlot[{ x^3 - 2 x x + 3 x y y - 2 y^3 == 0,
x^3 + 3 x x y + 3 x y y + y^3 2 Sqrt@2 == 0},
{x, -2, 2}, {y, -2,  2}]


Solving it:

N[Solve[x^3 + 3 x x y + 3 x y y + y^3 2 Sqrt@2 == 0 &&
x^3 - 2 x x + 3 x y y - 2 y^3 == 0, {x, y}, Reals]]

(*{{x -> 0., y -> 0.}, {x -> 1.11777, y -> -0.502861}}*)
`
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