Volume of 4-cube and n-cubes with edge length 3cm What is the volume (I have to clarify this: not the "volume" in the context of a 4-cube but in our everyday life where it is always a cubic unit) of a $4$-cube in $\text{cm}^3$? A 4-cube contains eight $3$-cubes. So it should be $8 * 3\text{cm}^3 = 216\text{cm}^3$. But is this all there is? 
And if so, what volume does a $5$-cube have? A 5-cube has besides some $4$-cubes, 40 $3$-cubes, which have $40 * 3\text{cm}^3 = 1080\text{cm}^3$ in total.
Is this the exact volume of a $5$-cube?
Also consider the following thought experiment: You have a $3$-cube and fill it with water. You pour it's content into a $5$-cube. How many times could you do this until the $5$-cube is full?
 A: Consider an analogy. A 1-cube is just a segment of length $L\ \textrm{cm}$. The 2-cube, the square, has four 1-cubes for its sides. Does this mean that the area of a square is $4L\ \textrm{cm}?$ No. The perimeter of the square is measured in centimeters, but its area is measured in square centimeters, and has nothing (directly) to do with the 1-cubes that are on its face.
In $n$ dimensions, volume has units $\textrm{cm}^n$. There is potential for confusion because we use the same word -- volume -- to refer to different quantities: just like length and area are not the same thing, 3-dimensional volume and 4-dimensional volume are also unrelated, and you don't calculate the latter by adding up the former. The $n$-cube of side length $L$ has $n$-dimensional volume $L^n\ \textrm{cm}^n$. Notice the units. Adding up the volumes of the $n-1$-cubes gives you the $n-1$-dimensional volume of the $n$-cube's boundary -- you can think of this as surface area -- which is $2nL^{n-1}\ \textrm{cm}^{n-1}.$
