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 $x^3+y^3 = (x+y)(x^2-xy+y^2)$. 

►If $z\equiv0\pmod7$ we have $x^3+y^3\equiv 3\pmod7$ but by  Fermat's little theorem $w^6\equiv1\pmod7$ if 
$w\not \equiv 0\pmod7$ then $w^3\equiv\pm1\pmod7$ so either $\pm1\pm1\not\equiv3\pmod7$ or $0\pm1\not\equiv\pmod7$.
►If $z\not\equiv0\pmod7$ we have the equation $x^3+y^3\equiv 1+3=4\pmod7$ and similar reasoning show the impossibility.
►If $x^3+y^3=z^6+3$ have solution this solution must satisfy the congruence modulo $7$ so there is not solutions.
