Prove that for any $x, a \in \mathbb{R} $ with $x\neq0$ the following statement holds:
$$x(x − 2a^2) > 0 ⇔ |x − a^2| > a^2$$
This is what I did but I'm not sure, and I think it is too long:
$$x(x - 2a^2) > 0$$
This inequality is true if and only if one of the following cases are satisfied:
Case 1: $x > 0$ and $x > 2a^2$
Case 2: $x < 0$ and $x < 2a^2$
Now, let's simplify the right-hand side of the statement:
$$\left\lvert x - a^2 \right\rvert > a^2$$
This inequality is true if and only if one of the following cases is satisfied:
Case 3: $x - a^2 > a^2$
Case 4: $x - a^2 < -a^2$
Let's consider each case:
Case 1: $x > 0$ and $x > 2a^2$
In this case, we have $x - a^2 > 0$ (since $a$ is a real number, $a^2$ is non-negative). Thus, $ \left\lvert x - a^2 \right\rvert = x - a^2$, and we can rewrite the right-hand side of the statement as follows: \begin{align*} x - a^2 &> a^2 \\ \implies x &> 2a^2 \end{align*}
This is equivalent to the left-hand side of the statement. Therefore, the statement holds for this case.
Case 2: $x < 0$ and $x < 2a^2$
In this case, we have $x - a^2 < 0$. Thus, $ \left\lvert x - a^2 \right\rvert = -(x - a^2)$, and we can rewrite the right-hand side of the statement as follows: \begin{align*} -(x - a^2) &> a^2 \\ \implies x &< 2a^2 \end{align*} This is the negation of the left-hand side of the statement. Therefore, the statement holds for this case as well.
Case 3: $x - a^2 > a^2$
In this case, we have $x > 2a^2$. Thus, the left-hand side of the statement is true, and the right-hand side is also true. Therefore, the statement holds for this case.
Case 4: $x - a^2 < -a^2$
In this case, we have $x < 0$. Thus, the left-hand side of the statement is false, and the right-hand side is also false. Therefore, the statement holds for this case as well.
Since the statement holds for all possible cases, it is true for any $x, a \in \mathbb{R}$ with $x \ne 0$.
I found I had to do a lot of extra thinking -- basically completing the proof -- in order to decide whether this set of cases actually proves the theorem. I think the cases are sufficient, but I don't think you've done enough to argue that they are.
There actually are not four cases here. Let's take a look at your cases:
Case 1: $x > 0$ and $x > 2a^2$
This case is really just $x > 2a^2$, because $2a^2 \geq 0$ for any $a$. So if $x > 2a^2$ then we know $x > 0$ already and the condition $x > 0$ is redundant.
Case 2: $x < 0$ and $x < 2a^2$
This case is really just $x < 0$, because $2a^2 \geq 0$ for any $a$. So if $x < 0$ then we know $x < 2a^2$ already and the condition $x < 2a^2$ is redundant.
Case 3: $x - a^2 > a^2$
Add $a^2$ to both sides, and we find that this case is simply $x > 2a^2.$ That is, it is Case 1 again.
Case 4: $x - a^2 < -a^2$
Add $a^2$ to both sides, and we find that this case is simply $x < 0.$ That is, it is Case 2 again.
Usually when one considers multiple cases in order to prove a statement using absolute values, one chooses an exhaustive set of cases. That is, the cases should cover all possible worlds in which the statement could be evaluated. An exception is that when proving the implication $P \implies Q,$ we get the case where $P$ is false "for free," because as long as $P$ is false we immediately know that $P \implies Q$ is true. In that case we say the implication is true vacuously, and I think people may often not even bother to describe this case (since it is so trivial) but will simply examine cases in which $P$ is true.
Considering your statement as a biconditional, $P \iff Q,$ we can't claim it is true vacuously when $P$ is false, so if we're just going to go through a set of cases one by one and verify the statement $ x(x − 2a^2) > 0 \iff \lvert x − a^2\rvert > a^2, $ you would want another case:
Case 5: $0 \leq x \leq 2a^2.$
In this case $x(x − 2a^2) > 0$ and $\lvert x − a^2\rvert > a^2$ are both false, so the statement to be proved is true.
But if you recast the statement $ x(x − 2a^2) > 0 \iff \lvert x − a^2\rvert > a^2 $ as two implications,
$$ (x(x − 2a^2) > 0 \implies \lvert x − a^2\rvert > a^2) \land (\lvert x − a^2\rvert > a^2 \implies x(x − 2a^2) > 0), $$
you can use Case 1 and Case 2 to prove the first implication, ignoring the vacuously true case, and then you can use Case 3 and Case 4 (that is, Case 1 and Case 2 again) to prove the second implication, again ignoring the vacuous case.
Personally, if I were going to argue this by case logic, I would use the three cases $x < 0,$ $0 \leq x \leq 2a^2,$ and $x > 2a^2,$ which are an exhaustive set of cases, and prove in each case that the two sides of the $\iff$ are either both true or both false.