The simplest way to explain arithmetic mean is in terms of "equal sharing":
Abe has 12 cookies, Brianna has 8 cookies, and Chuck has 7 cookies. If they were to redistribute them so that they all have the same amount, how many would each get?
Obviously the way you answer this question is to find the total amount of cookies ($12 + 8 + 7 = 27$) and then divide the cookies among the three people ($27/3 = 9$). That's precisely what the computation of arithmetic mean does.
Is that what you're looking for?
Edited to add:
Here's another viewpoint that might help. We would like to find some number $N$ that is in the "middle" of the set $ \{12, 8, 7 \}$ (using the same numbers from the example above). What does "in the middle" mean? Well, one way to interpret this vague phrase is to imagine that we already had such an $N$ in hand, and we compute the three quantities $12-N, 8-N,7-N$. These three quantities tell us how far $N$ is from each of three pieces of information -- call these the "deviations".
What if we made a bad choice of $N$? For example, if each of the three deviations were positive, then that would mean that $N$ is smaller than each of the three original numbers, which we don't want. If each of the three deviations were negative, then that would mean that $N$ is larger than each of the three original numbers -- again bad. For $N$ to be in the middle, we would want some of the deviations to be positive and some of them to be negative. In fact, if we could choose $N$ so that the positive deviations exactly cancel out the negative deviations, then we will feel like we've really found the "middle".
Let's translate that now into a computation. We want to find $N$ such that
$$(12-N) + (8-N) + (7-N) = 0$$
If you now think about what it would take to solve this equation, you will quickly realize that you end up adding the three numbers in your dataset together and then dividing by 3.
$(A-B)(A-C)\;=\;A^2-AB-AC+BC\;=\;A(A-(B+C))+BC.$
Therefore, with $A=10^2$ and $B=3$ and $C=4$, we have $$(97)(96)=(10^2-3)(10^2-4)=10^2(10^2-(3+4))+3\cdot 4= (100)(93) +12.$$
Another example: $(91)(92)=10^2(10^2-(9+8))+9\cdot8=8300 +72.$
Best Answer
This is really just ordinary long multiplication, but with lazy evaluation of carries, and the rows are inverted. That is, imagine multiplying $255 \times 25$ the usual way. The first row would have the result of multiplying $255 \times 5$, which is $1275$, and the second row would have the result of multiplying $255 \times 20$, which is $5100$. We then add $1275+5100 = 6375$.
Well, this is the same thing, except that first of all, we reverse the rows, so that we multiply $255 \times 20$ first. Secondly, we leave all the carries unevaluated at first, so that instead of writing $5100$, we write ${}^04^10^10^00$, where the $1$'s are the carry results of multiplying $5 \times 2$. The first $1$ will carry leftward to join the $4$, and the second $1$ will carry leftward to join the $0$, but unlike traditional multiplication, we leave that undone for the time being.
The second row is then the result of $255 \times 5$, except that instead of $1275$, we have $^10^25^25$. Now, finally, we add ${}^04^10^10^00+{}^10^25^25$:
\begin{align} ^04^10^10^00 & \\ \underline{+\text{ } ^10^25^25} & \\ 6^\phantom{0}3^\phantom{0}7^\phantom{0}5 \end{align}
where now all the carries are done at once: $4+1+1 = 6, 1+2 = 3, 5+2 = 7, 5 = 5$. The lattice lines are drawn so as to visually facilitate this, but otherwise, there's nothing terribly magical going on here. Notice that there are no carries in the addition; otherwise, this method would not appear to as great an advantage. I consider that a bit of intellectual dishonesty, really.
ETA: I suppose that if the purpose was to teach the method, rather than to advocate it, then making the example more straightforward is all right.