Give a combinatorial proof of the following identity: $n^3 = n(n - 1)(n - 2) + 3n(n - 1) + n$

Question

Give a combinatorial proof of the following identity: Let $n$ be a positive integer. $n^3 = n(n - 1)(n - 2) + 3n(n - 1) + n$.

How would I go about proving this identity combinatorially? Thank you.

Answer 1

$n^3$ is the number of possible triplets of numbers from $1$ to $n$. $n(n-1)(n-2)$ is the number of triplets where no $2$ numbers are the same. Then, $n(n-1)$ is the number of triplets with $(a, a, b)$ ($a\not =b$), so $3n(n-1)$ is the number of triplets with exactly two equal elements. $n$ is the number of triplets with all elements the same.

Answer 2

Consider an $n\times n\times n$ cube. We partition it into five pieces:

First, there is a large $n\times (n-1)\times (n-2)$ block. This leaves an L-shaped piece, of height $n$, with one arm width $1$ and the other arm of width $2$. We partition the arms into three slabs, each of size $n\times 1\times (n-1)$. This leaves the corner of the L, which is a single $n\times 1\times 1$ long piece.

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