Find all possible remainders. 
Suppose $b\in\mathbb Z$. Find all possible remainders of $b^3$ divided by $7$. 

I know $b^6=1\bmod7$ which means $(b^2)^3=1\bmod7$ so all square numbers^3 leave $1$ as a remainder, but how to continue?
 A: Notice that this depends only on the value of $b$ mod $7$, since $(b+7k)^3 = b^3 + (\text{multiple of 7})$. So we can work wlog with $b = 0, 1, \dots, 6$.
Your method is also faulty: just because $b^6 \equiv 1 \pmod{7}$, that doesn't mean $b^2 \equiv 1 \pmod{7}$. For example, $3^2 \equiv 2 \pmod{7}$.

There's a really brute-force way to do it, modulo 7. I'll use $=$ instead of $\equiv$ to save a bit of typing:
$0^3 = 0, 1^3 = 1, 2^3 = 8 = 1, 3^3 = 27 = -1, 4^3 = 64 = 1, 5^3 = (-2)^3 = -1, 6^3 = (-1)^3 = -1$
yielding the answers $0, 1, -1$.
Really, you only need to check $0, 1, 2, 3$ (and then negate those answers), because $(7-3)^3 \equiv (-3)^3 = - (3^3)$.
A: It doesn't matter if they ask you about $b^3$ or about $b^5$ or about say $b^5+b^2+11$. The method is always the same: if your modulo is N, you just take the numbers $0, 1, 2, ..., N$ and you just put them in the formula (for $b$), and you observe what the outputs (modulo N) are.   
The reason for this is the validity of this statement.
If $a\equiv b\pmod N$ and if $f$ is any polynomial, then $f(a)\equiv f(b)\pmod N$.
This statement is easily provable using the basic properties of congruences.    
