Find all solutions of the system: (hint: factor the left 
side of the second equation: 

x-y = 3
x^3 - y^3 = 387 

So now after factoring I have this new system: 
 x-y = 3
(x-y)(x² + xy + y²) = 387 

but now how do I continue the problem? 
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Since you have x-y = 3 you can substitute 3 for (x-y) in

          (x-y)(x² + xy + y²) = 387 

getting

              3(x² + xy + y²) = 387

Divide both sides by 3

                 x² + xy + y² = 129

Now since x-y = 3, solving for y gives

                            y = x-3

So substitute (x-3) for y in

                 x² + xy + y² = 129

         x² + x(x-3) + (x-3)² = 129

    x² + x² - 3x + (x-3)(x-3) = 129

         2x² - 3x + (x²-6x+9) = 129

       2x² - 3x + x² - 6x + 9 = 129

               3x² - 9x - 120 = 0 

              3(x² - 3x - 40) = 0 

              3(x - 8)(x + 5) = 0

That gives two possible values for x:

             x = 8 and x = -5

Now substituting x = 8 into
                           
                            y = x - 3
                            y = 8 - 3
                            y = 5

So one solution is (x, y) = (8, 5)  

Substituting x = -5 into
                           
                            y = x - 3
                            y = -5 - 3
                            y = -8

So the other solution is (x, y) = (-5, -8) 

Edwin