Can't argue with success? Looking for "bad math" that "gets away with it" I'm looking for cases of invalid math operations producing (in spite of it all) correct results (aka "every math teacher's nightmare").

One example would be "cancelling" the 6's in

$$\frac{64}{16}.$$

Another one would be something like

$$\frac{9}{2} - \frac{25}{10} = \frac{9 - 25}{2 - 10} = \frac{-16}{-8} = 2 \;\;.$$

Yet another one would be

$$x^1 - 1^0 = (x - 1)^{(1 - 0)} = x - 1\;\;.$$

Note that I am specifically not interested in mathematical fallacies (aka spurious proofs).  Such fallacies produce shockingly wrong ends by (seemingly) valid means, whereas what I am looking for all cases where one arrives at valid ends by (shockingly) wrong means.

Edit: fixed typo in last example.
I was quite amused when a student produced the following when cancelling a fraction:

$$\frac{x^2-y^2}{x-y}$$

He began by "cancelling" the $x$ and the $y$ on top and bottom, to get:

$$\frac{x-y}{-}$$

and then concluded that "two negatives make a positive", so the final answer has to be $x+y$.
Well, using $\frac{dy}{dx}$ as a ratio is also one such example. It carries no meaning at all to use  $\frac{dy}{dx}$ as ratio but it works each time to get you to the right result instead of appreciating the limit process behind it.
I think it is as much abuse of notation as is eliminating 6's from a fraction, it is just that in this case the notation is designed such that each time you abuse it, you will get the right answer but you won't know actually why if you just rely on the notation. For example:
$\frac{dy}{dx}=\frac{dy}{dt} \frac{dt}{dx}$ due to chain rule and not due to treating $\frac{dy}{dx}$ as a fraction just like $\frac{x^2-y^2}{x-y}=x+y$ by algebra and not by 

$$\frac{x^2-y^2}{x-y}$$

"cancelling" the $x$ and the $y$ on top and bottom, to get:

$$\frac{x-y}{-}$$

and then conclude that "two negatives make a positive", so the final answer is be $x+y$.

Now if you see a student prove $\frac{5^2-3^2}{5-3}=8$ by the wrong method, will he be correct as this method of proof will hold for all $x,y \in R$ until $x \neq y$.

Also in three dimensions you have stuff like $ \frac{\partial x}{\partial y}\cdot\frac{\partial y}{\partial z}\cdot\frac{\partial z}{\partial x}=-1. $

Also, $dy$ isn't even defined , let alone be manipulated like a real number.