$1000$ small cubes are assembled into a larger cube. If one layer of small cubes is removed, how many remain? 
If you have $1000$ small cubes, which are $1$cm $\times$ $1$cm $\times$ $1$cm, and you build
  a bigger cube with a volume of $1000$cm$^3$, and then you remove the outer layer, how many cubes are left?

Now I know there are 10 small cubes on each side, as the volume is $1000$cm$^3$. however once I remove the 'outer layer' how do I 'figure' out how many cubes are left? 
I know the answer is actually $8^3 = 512$, but I cannot seem to logically get there, what is confusing is also the edges share the same cube. 
Can anyone shed some light? 
Thank you. 
 A: If you remove the outer layer, it means every single side became shorter by exactly $2$.
So you have a cube with a side of $8$ in hand, and its volume is indeed
$$8^3 = 512.$$
A: The other answer is very short, elegant and definitely the best way to do it. But it requires that you happen upon a small insight. In case that fails, here is a more straight-forward approach. It is a bit longer, and takes a tiny bit more work, but it's the solution you find if you just sit down and start counting.
How many cubes were removed? There are six faces, and each face has $100$ cubes, so the first naive answer would be $6\cdot 100$. However, as you say, it's not quite as simple as there are cubes shared between faces. The standard way to correct this is the so-called inclusion-exclusion principle. We start with our $600$ cubes, and then subtract every edge (since each edge is counted twice). There are 12 edges, and each edge has $10$ cubes, so we get
$$
600-12\cdot 10 = 480
$$
However, this isn't entirely correct either, as we have the corners. Each corner was counted once for each face it's in (three times), then subtracted once for each edge it was in (three times), meaning we haven't really included them in our count yet. So we add them back. Thus the final tally is that we removed a total of
$$
600-120 + 8 = 488
$$
cubes, meaning we have $1000-488 = 512$ left.
Alternatively, we could take more care in our counting: Each face has $8\cdot 8$ cubes that it doesn't share with any other face, each edge has $8$ cubes it doesn't share with any other edges, and there are 8 corners. This also gives us a total of $6\cdot 64 + 12\cdot 8 + 8 = 488$ cubes removed.
A: Using the identity $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$ and substituting $a=10$ and $b=2$:
$$(10-2)^3 = 10^3 - (3\times10^2\times2) + (3\times10\times2^2) - 2^3 = 1000 - 600 + 120 - 8 = 512$$
Each term in the above has a geometric significance:


*

*$1000$ is the number of small cubes in the original cube.

*Each face of the original cube contains $10\times10=100$ small cubes, so the effect of removing the small cubes on all six faces, before allowing for double or triple counting, is a deduction of $600$.

*Adding back $120$ allows for double counting of the $10$ small cubes on each of the twelve edges.

*Finally, deducting $8$ is necessary because the small cubes at the eight vertices have been triple counted within the deduction of $600$ but also triple counted within the adding back of the $120$, so need to be deducted once more.
