The quotient set $\mathbb{Z}/{\sim}$ where $a\sim b \iff$ $a^3 \equiv b^3 \pmod{7}$ Consider equivalence relation $\sim$ on $\mathbb{Z}$ s.t. $\forall a,b\in \mathbb{Z}$:
$$a\sim b \iff a^3 \equiv b^3\!\!\!\! \pmod{7}.$$
There are some questions, but I am struggling with: determine the quotient set $\mathbb{Z}/\!\sim$
I have tried working with:
$7|a^3-b^3=(a-b)(a^2+ab+b^3) \,\,\,$so$\,\,\, 7|a-b\,\,\,$or$\,\,\,7|a^2+ab+b^2$
but without results.
 A: Numbers that are congruent modulo $7$ will obviously be related by $\sim$ and therefore also in the same equivalence class.
So all you need to do is figure out which of the congruence classes modulo $7$ combine to form each $\sim$-equivalence classes.
Instead of trying to find a general algebraic argument that does everything at once, simply do it case-by-case by computing $a^3\mod 7$ for each possible congruence class. See which of them give the same result. Job done.
A: Note that $7$ is prime, and $3$ divides $p-1$. That means that the only cubes mod $7$ are $0$ and two others (two, because $(7-1)/3=2$). Clearly $1$ and $-1\equiv6$ are cubes, with $1$ and $-1$ being some cube roots of those cubes.
So the quotient set has three members. The classes of $0$, $1$, and $6$. The only other things in the class of $0$ are other numbers that are $0$ mod $7$. There are other things in the class of $1$ and other things in the class of $-1$, and has been pointed out by others, direct examination of the congruence classes mod $7$ is an efficient way to identify them.
