Prove that there are exactly two solutions to the equation $x^3 = x^2$. This is Velleman's exercise 3.6.12.c:
Prove that there are exactly two solutions to the equation $x^3 = x^2$.
Here's my proof of it:
Proof. 
Existence. Let $x = 0$ then clearly $x^3 = x^2$ and let $x = 1$ then clearly $x^3 = x^2$.
Uniqueness. We choose an arbitrary $z$ such that $z^3 = z^2$. Now we consider two exhaustive cases:
Case 1. $z = 0 = x$, then clearly $0 = 0$.
Case 2. $z \neq 0$, then $z^2 \neq 0$ and dividing the equation $z^3 = z^2$ by $z^2$, we get $z = 1 = x$.
Is my proof correct?
Thanks. 

Yes, your proof is correct.

But here's a more standard proof . .  .

\begin{align*}
&x^3=x^2\\[4pt]
\iff\;&x^3-x^2=0\\[4pt]
\iff\;&x^2(x-1)=0\\[4pt]
\iff\;&x^2=0\;\;\text{or}\;\;x-1=0\\[4pt]
\iff\;&x=0\;\;\text{or}\;\;x=1\\[4pt]
\end{align*}
