If $(x-a)^2=(x+a)^2$ for all values of $x$, then what is the value of $a$? If $(x-a)^2=(x+a)^2$ for all values of $x$, then what is the value of $a$?
At the end when you get $4ax=0$, can I divide by $4x$ to cancel out $4$ and $x$?

$$(x-a)^2 = (x+a)^2$$
$$x^2 - 2ax + a^2 = x^2 + 2ax + a^2$$
$$x^2 - 2ax + a^2 = x^2 + 2ax + a^2$$
$$-2ax = 2ax$$
$$-a = a$$
Note that this statement is only true when $a=0$, which is thus your solution.
