Solve for integers $x, y, z$ such that $x + y = 1 - z$ and $x^3 + y^3 = 1 - z^2$. 
Solve for integers $x, y, z$ such that $x + y = 1 - z$ and $x^3 + y^3 = 1 - z^2$.

I think we'll have to use number theory to do it. Simply solving the equations won't do.
If we divide the second equation by the first, we get:
$$x^2 - xy + y^2 = 1 + z.$$
Also, since they are integers $z^2 \ge z \implies -z^2 \le -z$. This implies
$$x + y = 1 - z \ge 1 - z^2 = x^3 + y^3.$$
This shows that atleast one of $x$ and $y$ is negative with the additive inverse of the negative being larger than that of the positive.
I have tried but am not able to proceed further. Can you help me with this?
Thanks.

$$y=-x, z=1$$
$${y = \frac{-1 + x - \sqrt{3} \sqrt{3 - 2 x - x^2}}{2}, z = \frac{3 - 3 x + \sqrt{3} \sqrt{3 - 2 x - x^2}}2}$$
$${y = \frac{-1 + x + \sqrt{3} \sqrt{3 - 2 x - x^2}}{2}, z = \frac{3 - 3 x - \sqrt{3} \sqrt{3 - 2 x - x^2}}2}$$
All integer solutions:
$x=-3,   y=-2,   z=6$
$x=-2,   y=-3,   z=6$
$x=-2,   y=0,   z=3$
$x=0,   y=-2,   z=3$
$x=0,   y=1,   z=0$
