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

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.

## closed as off-topic by Morgan Rodgers, José Carlos Santos, Archer, Cesareo, Parcly TaxelMar 10 at 3:39

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$$n^3$$ 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.

• I would assume that the second sentence needs some editing. – user Mar 6 at 20:36
• Thank you, this helps tremendously. But how did you determine that $3n(n-1)$ is the set of all lists with exactly two equal elements? – M Lee Mar 6 at 20:46
• Because, either first equals the second, or second equals the third or third equals the first. – enedil Mar 6 at 20:47
• "that 3n(n−1) is the set of all lists with exactly two equal elements?" Because there are $n(n-1)$ ways you can pick which two elements, and there are $3$ ways you can arrange the two like elements and the one unlike element. – fleablood Mar 6 at 22:00

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.

MS Paint masterpiece below:

• How is it combinatorial? – enedil Mar 6 at 20:11
• We are double-counting the same object. If you like, take the cube as a collection of $n^3$ small cubes, and we are counting the number of those small cubes. – vadim123 Mar 6 at 20:11
• Ok, fair enough – enedil Mar 6 at 20:12