Simple mod 7 problem I need to Show that $7x^3 + 2 = y^3$ has no solutions in integers x and y.
The solution I am given is: Suppose there are solutions to this equation.
Then mod 7 we have $2 ≡ y^3$ (mod $7$) and hence $2$ is a cube mod $7$. However the cubes mod $7$ are $0$ and plus/minus $1$.
What does $2$ is a cube mod $7$ mean? Also, why are $0$ and plus or minus $1$ cubes mod $7$?
I have tried to cube the residues mod 7 i.e {$1^3,2^3,3^3,4^3,5^3,6^3$} and I get {$1,1,6,1,6,6$}.
Where have I got the wrong idea here?

${\rm mod}\ n\!:\,\ a\,$ is a cube if $\  a \equiv b^3\,$ for some integer $b$.
Note $\,{\rm mod}\ 7\!:\ y^3\equiv 2\,\overset{\rm square}\Rightarrow\, y^6\equiv 4\,$ contra little Fermat. 
Alternatively cubes are $\,\equiv \{0,\pm1,\pm2,\pm3\}^3\equiv \{0,\pm1\},\,$ but $\,2\not\equiv 0,\pm1\,$ so $\,2\,$ is not a cube.
