Proving no integer solution for $x^3+y^3+4=z^3$ I've been trying to shoe that there are no integer solutions for the equation:
$$x^3+y^3+4=z^3$$
What I've tried out so far is to prove by contradiction and to get to a contradiction using modulus.
Let us assume that there are $x,y,z\in\mathbb{Z}$ so that they solve the equation above. therefore:
$$x^3+y^3+4=x^3+3x^2y+3y^2x+y^2-3x^2y-3y^2x+4=(x+y)^3-3xy(x+y)+4=z^3$$
$$4-3xy(x+y)=z^3-(x+y)^3$$
What I did next is I checked all the possibilities for $x,y,z$ to be odd/even: $x,y,z\equiv 0 \mod2$, $x,z\equiv 0 \mod2, y\equiv 1 \mod2$ and so on...
All cases give a contradiction except for the case where $x,y$ are odd and $z$ is even.
Help?

Hint: Try the equation mod $9$. The cubes mod $9$ are $0,\pm1$.
Solution:

 Write the equation as $x^3+y^3+(-z)^3=-4$. The range of the LHS mod $9$ is $-3,\dots,3$ and so is never $-4$.

