, 2015

We subtract the equations to get
$$x^3 - y^3 + x^2 + y^2 + xy + 2x - 2y + 2 = 0.$$
We rearrange the left side
$$\begin{align*} x^3 - y^3 + x^2 + y^2 + xy + 2x - 2y + 2 &= \\ = (x - y)(x^2 + xy + y^2) + (x^2 + xy + y^2) + 2(x - y + 1) = \\ = (x - y + 1)(x^2 + xy + y^2) + 2(x - y + 1) = \\ = (x - y + 1)(x^2 + xy + y^2 + 2). \end{align*}$$
Since $x^2 + xy + y^2 + 2 = \frac{1}{2}(x^2 + (x+y)^2 + y^2) + 2 > 0$ we must have $x - y + 1 = 0$. The value of the expression $x - y$ equals $-1$.

From the second equation we express $x = y^3 - y^2 + 3y$, insert it in the second equation, and simplify to get
$$y^{9} - 3y^{8} + 12y^{7} - 18y^{6} + 34y^{5} - 19y^{4} + 21y^{3} + 11y^{2} + 4y + 2 = 0.$$
With quite a lot of skill we can factor the left side of the equation to
$$(y^{3} - y^{2} + 2y + 1)(y^{6} - 2y^{5} + 8y^{4} - 7y^{3} + 13y^{2} + 2) = 0.$$
Since
$$\begin{align*} y^{6} - 2y^{5} + 8y^{4} - 7y^{3} + 13y^{2} + 2 &= y^{4}(y^{2} - 2y + 1) + 7y^{2}\left(y^{2} - y + \frac{1}{4}\right) + \frac{45}{4}y^{2} + 2 = \\ = y^{4}(y - 1)^{2} + 7y^{2}\left(y - \frac{1}{2}\right)^{2} + \frac{45}{4}y^{2} + 2 > 0, \end{align*}$$
we must have $y^3 - y^2 + 2y + 1 = 0$ which gives $y = y^3 - y^2 + 3y + 1$. Therefore
$$x - y = (y^3 - y^2 + 3y) - (y^3 - y^2 + 3y + 1) = -1.$$