$x + y = 7 $and $x^3 - y^3 = 37$, find $xy$ $x+y=7$
  $x^3-y^3 = 37$
  
  Find $xy$


I have tackled this one but I am stuck, when I get to this point:
$$(x-y)((x+2)^2 -xy) = 37$$
I would be glad if you could give me any suggestions.
Setting $y=7-x$, we obtain
$$
0=x^3-y^3-37=(2x^2 - 13x + 95)(x - 4).
$$
For integer solutions, the quadratic term has no integer root, so that $x=4$ and $y=3$, hence $xy=12$. Over the real numbers, this is also the unique solution, since $2x^2-13x+95=0$ has two complex, non-real solutions.
$(49-4xy)(49-xy)^2=37^2$
, $37$ is prime. So $49-4xy=1$ then
$$xy=12$$