Let's ignore the cans for a moment.
Suppose you have a box of dimensions: $\text{Height: 60 mm, Length: 100 mm, Width: 100 mm}$ $\text{Volume = 60 * 100 * 100 = 600,000 mm}^3$
Now, suppose we cut off 1 mm of width from this box. When we do this, we're not just losing width, but in fact, we're losing volume. This is because we're really cutting the box into two (unequal) smaller boxes. These two smaller boxes have dimensions:
$\text{Height: 60 mm, Length: 100 mm, Width: 99 mm}$
$\text{(Volume = 60 * 100 * 99 = 594,000 mm$^3$)}$
and
$\text{Height: 60 mm, Length: 100 mm, Width: 1 mm}$
$\text{(Volume = 60 * 100 * 1 = 6,000 mm$^3$).}$
This is where the 6,000 mm$^3$ comes from: $600,000 - 594,000 = 6,000.$
As anon mentioned in the comments, here is another way of seeing this is by using the distributive law: $\text{60*100*100 - 60*100*99 = 60*100*(100 - 99) = 60*100*1 = 6,000.}$
Edit: Let's look at the second example. We have:
$\text{22*33*44 = 31,944}$
$\text{19*20*21 = 7,980}$
Now, if we want to compare the lengths, widths, heights, surface areas, or volumes of the boxes, then we probably shouldn't multiply or divide the dimensions of different boxes. That is, the number 31,944 comes from the first box, whereas 20 and 21 come from the second box, so it doesn't make sense to divide $31,944$ by $21$ by $20$, because this doesn't tell us anything helpful.
Similarly, there is no reason to expect that the sum of the length, width, and height (22 + 33 + 44, and 19 + 20 + 21) will have anything to do with each other. I know it seems like there's a "missing 39 mm," but this 39 mm does not represent anything meaningful, and does not really have anything to do with (22 + 33 + 44 - 19 - 20 - 21).
Anyway, if this still doesn't answer your question, maybe thinking about this will help:
$\begin{align*} 31,944 & = 22^*33^*44 \\ & = (19 + 3)^*(20 + 13)^*(21 + 23) \\ & = 19^*20^*21 + (19^*20^*23 + 19^*13^*21 + 19^*13^*23 \\ & \ \ \ \ + 3^*20^*21 + 3^*20^*23 + 3^*13^*21 + 3^*13^*23) \\ & = 7,980 + 23,964 \end{align*}$