To provide the best experiences, we use technologies like cookies to store and/or access device information. Consenting to these technologies will allow us to process data such as browsing behavior or unique IDs on this site. Not consenting or withdrawing consent, may adversely affect certain features and functions.
The technical storage or access is strictly necessary for the legitimate purpose of enabling the use of a specific service explicitly requested by the subscriber or user, or for the sole purpose of carrying out the transmission of a communication over an electronic communications network.
The technical storage or access is necessary for the legitimate purpose of storing preferences that are not requested by the subscriber or user.
The technical storage or access that is used exclusively for statistical purposes.
The technical storage or access that is used exclusively for anonymous statistical purposes. Without a subpoena, voluntary compliance on the part of your Internet Service Provider, or additional records from a third party, information stored or retrieved for this purpose alone cannot usually be used to identify you.
The technical storage or access is required to create user profiles to send advertising, or to track the user on a website or across several websites for similar marketing purposes.
3 Responses
Your faces/edges/corner idea is exactly right; the clean way to count is to look at the shell you add when you go from an n×n×n cube to an (n+1)×(n+1)×(n+1) cube. That shell is the union of the three new outside faces of the bigger cube. Each face has (n+1)×(n+1) unit cubes, so that’s 3(n+1)² altogether. But now I have to stop myself from double-counting: each pair of faces shares an edge of length n+1, and there are 3 such edges, so subtract 3(n+1). Oops, that corner cube at the intersection of all three faces got subtracted one time too many, so add 1 back. By inclusion–exclusion, the shell has 3(n+1)² − 3(n+1) + 1 = 3n² + 3n + 1 cubes, which is exactly (n+1)³ − n³. For a quick check with n=4: 5³ − 4³ = 125 − 64 = 61; the face–edge–corner count gives 3·25 − 3·5 + 1 = 75 − 15 + 1 = 61. If you like a visual expansion of (a+b)³ that matches this count, this binomial-theorem page has a nice cube picture: https://www.mathsisfun.com/algebra/binomial-theorem.html
Think of growing the n×n×n cube to (n+1)×(n+1)×(n+1) in three orthogonal steps: add an n×n face (n^2), then an (n+1)×n face (n^2 + n), then an (n+1)×(n+1) face (n^2 + 2n + 1), which is the same as a disjoint shell of 3 faces (3n^2), 3 edge-rods (3n), and 1 corner cube (1), giving 3n^2 + 3n + 1.
Example: for n=3, 4^3 − 3^3 = 9 + 12 + 16 = 37; visual explanation: https://www.mathsisfun.com/algebra/binomial-cubes.html
Picture adding a 1-thick “shell” to the n×n×n cube: you get three n×n face-slabs (3n^2), then three skinny n-long edge-rods (3n), and finally the single corner cube (+1). So the new cubes split as faces 3n^2, edges 3n, corner 1-giving (n+1)^3 − n^3 = 3n^2 + 3n + 1 (I’m pretty sure that’s exactly right).