Solving the diophantine equation $x^3+y^3 = z^6+3$ I've the following problem:

Show that the congruence $x^3+y^3 \equiv  z^6+3\pmod{7}$ has no solutions. Hence find all integer solutions if any to $x^3+y^3-z^6-3 = 0.$

We can rearrange the first equation to $z^6 \equiv (x^3+y^3-3) \mod{7}$. But $z^6\equiv 1\mod{7}$ so this is only possible when $x^3+y^3=4$ which no $x,y \in  \mathbb{Z}$ satisfy. 
Now I don't know if that helps at all solve the second bit. We have $z^6 = x^3+y^3-3$. I know that $x^3+y^3 = (x+y)(x^2-xy+y^2)$. 

Any solution for $x^3+y^3-z^6-3=0$ is also a solution mod $7$. Therefore there are no solutions to the original equation.
