Show that $x^3 \equiv 0,1,8 \mod{9}$ (Critique my solution) My solution is based on the idea that it is sufficient to consider the cubes less than 9. Ie $0^3 \equiv 0, 1^3\equiv 1, 2^3 \equiv 8$. Since these are the only cubes in the residue class$\mod{9}$ these are the only possibilities for $x^3 \mod{9}$
Is this a sufficient motivation? 
New solution: 
By computation, $0^3 \equiv 3^3 \equiv 6^3 \equiv 0 \mod{9}$ and $ 1^3 \equiv 4^3 \equiv 7^3 \equiv 1 \mod{9}$ and $ 2^3 \equiv 5^3 \equiv 8^3 \equiv 8 \mod{9}$. For any $x$, we can can calculate $x^3\mod{9}$ by reducing $x \equiv y \mod{9}$ where $0 \leq y \leq 8$ , and so $x^3 \equiv y^3 \mod 9$  where $y^3$ is one of our cases above, ie either 0,1 or 8.  

According to that reasoning, since $1^3\equiv1,2^3\equiv8,3^3\equiv0\pmod{27}$, we have that $x^3\equiv0,1,8\pmod{27}$ for any $x$. But $4^3\equiv10\pmod{27}$.
