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.

The problem : solve the equation $$e^{(x-2)}+e^{(x+8)}=e^{(4-x)}+e^{(3x+2)}$$
Incorrect method
As everybody knows $$e^a+e^b=e^{ab}$$
So let's apply it to solve the equation :
$\begin{array}{lll}
e^{(x-2)}+e^{(x+8)}=e^{(4-x)}+e^{(3x+2)} & \iff & e^{(x-2)(x+8)}=e^{(4-x)(3x+2)} \\
                                 & \iff & (x-2)(x+8)=(4-x)(3x+2) \\
                                 & \iff & x^2+6x-16=10x-3x^2+8 \\
                                 & \iff & 4x^2-4x-24=0 \\
                                 & \iff & x^2-x-6=0 \\
                                 & \iff & (x+2)(x-3)=0 \\
\end{array}$
Thus $x=-2$ or $x=3$.
Correct method:
Let's set $X=e^x$ and $a=e^2$ and substitute
$\frac Xa+a^4X=\frac {a^2}X+aX^3\iff X^2+a^5X^2=a^3+a^2X^4\iff(a^2X^2-1)(X^2-a^3)=0$
Leading also to $x=-2$ and $x=3$.



The problem : simplify the expression $\sin(x+y)\sin(x-y)$
Incorrect method
As everybody knows $$f(a+b)f(a-b)=f(a)^2-f(b)^2$$
Applying the formula gives immediately 
$\sin(x+y)\sin(x-y)=\sin^2(x)-\sin^2(y)$
Correct method
$\begin{array}{lll}
\sin(x+y)\sin(x−y) &=& \big[\sin(x)\cos(y)+\cos(x)\sin(y)\big]\ \big[\sin(x)\cos(y)−\cos(x)\sin(y)\big]\\
&=&\sin^2(x)\cos^2(y)−\cos^2(x)\sin^2(y)\\
&=&\sin^2(x)(1−\sin^2(y))−(1−\sin^2(x))\sin^2(y)\\
\require{cancel}&=&\sin^2(x)−\cancel{\sin^2(x)\sin^2(y)}−\sin^2(y)+\cancel{\sin^2(x)\sin^2(y)}\\
&=&\sin^2(x)−\sin^2(y)\\
\end{array}$
