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. 
 A: 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*}
A: Prove that there are exactly two solutions to the equation $x^3=x^2$
Here's two alternate approaches to this problem:
1) 
We can use a graphing calculator or computer program to plot the functions $y=x^3$ and $y=x^2$ as individual functions. Provided that we adjust the viewing window correctly, we can see that there are exactly two intersections between these two graphs. Which implies that there are 2 values for $x$ that satisfy the original equation $x^3=x^2$
2) (if we were asked this problem in a calculus class)
We can move all the $x$ terms to one side of the equations, resulting in the following: $0=x^3 - x^2$. We could then use the intermediate value theorem, to prove that there are two roots to the equation $0=x^3 - x^2$.
Note that this particular approach is slightly more cumbersome, as we would have to determine two intervals on which the IVT can be applied to prove the existence of a root. That said, this would prove nonetheless that exactly two solutions exist for $x^3=x^2$
Solving this algebraically is clearly the most efficient means to a solution, but in spirit of making things more complex than they need to be, I thought I'd share these alternate approaches for anyone who may be struggling to recognize the algebraic solution to this problem.
