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?

You basically have the right idea. It is just that you forgot two details: 


*

*modulo $7$ you have that $6$ and $-1$ are in the same class, so your $6$ is in fact just the $-1$. 

*You forgot the class $0$.
If you take this into account your calculation shows that the only cubes are $0,1,-1$ as claimed. 
In any case you see that (the class of) $2$ is not the cube of any class modulo $7$.
However, $2$ would have to be a cube if there were a solution to your original equation. Hence, there can not be a  solution. 
