Another way to proof $A^2=B^2$ When we have an eqaution in the form of $A^2=B^2$ the way that I've been taught would be to square both sides of the equation $\sqrt{A^2} = \sqrt{B^2}$ and the result would be that
$A = B$ and $A = -B$ , but why is this the case? $A = B$ seems obvious enough since it "cancels each other out", but we are not allowed to have $-B$ as a result of a $\sqrt x$ sign isn't that right? Because it is supposed to be reserved for only positive numbers right? Or are we allowed to do it because $(-B)^2$ results in $B^2$ and therefore is allowed? I am asking this because we have been taught to do it, but not the reasoning behind it and I would probably make careless mistakes because of it.
 A: A better way to look at this is as follows. If you know that $A^2=B^2$, then what can you say about $A$ and $B$? You know that one of the following is true, either $A=B$ or $A=-B$. You also do not know anything more about $A$ and $B$ than that.
You don't need to think about things cancelling each other out, or about taking square roots of anything, just think of it as a logical proposition.
If $A^2=B^2$ then the only possible logical explanation is that either $A=B$ or $A=-B$. "If your cat had kittens then your cat must be a female." "If your car stops running either you've run out of petrol or the engine is broken". You don't need to keep on tormenting numbers and equations and minus signs like you are doing.
Just to extend this, what do we know about $A$ and $B$ if $A^3=B^3$?

 We know that $A=B$. If $A=-B$ then $A^3=-B^3$.

A: I think it might be useful to look into a concrete example: find the possible $B$ that would satisfy
$$9 = B^2$$
One case is that $B$ is non-negative, and that corresponds to your "cancel each other out" case:
$$\begin{align*}
\sqrt 9 &= \sqrt{B^2}\\
3&=B
\end{align*}$$
Then another case is that $B$ is negative. You are right that the result of $\sqrt{\cdot}$ is non-negative, then if you apply $\sqrt{\phantom{B^2}}$ to $B^2$, the result has to be positive. But among $-B$ and $B$, here $-B$ is the positive number, and $\sqrt{B^2} = -B$:
$$\begin{align*}
\sqrt 9 &= \sqrt{B^2}\\
3 &= -B\\
B &= -3
\end{align*}$$
So $3$ and $-3$ are the only two possible $B$s that satisfy $9=B^2$.
Or in your question, either $A=B$ or $A=-B$.
A: If you want, there's no need for radicals. $A^2 = B^2$ implies $A^2 - B^2 = 0$ and therefore, $(A - B)(A + B) = 0$. So one of the equations $A - B = 0$ or $A + B = 0$ must be true. Thus $A = B$ or $A = -B$.
