Since Borsuk conjecture hold for centrally symmetric convex sets in $ \mathbb{R}^n$ so we can cut a hypercube into at least $ n+1$ disjoint parts.

Is there a method how can one do that?

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# Tag: cube’s

## Is there a method to cut a hypercube into disjoint cubes

## Products and sum of cubes in Fibonacci

## Numbers that can be written as a sum of three cubes in exactly one way (a^3 + b^3 + c^3)

## How to find the vole of a cube remaining after drilling a cylinder with a diameter larger than the cube’s side length?

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Since Borsuk conjecture hold for centrally symmetric convex sets in $ \mathbb{R}^n$ so we can cut a hypercube into at least $ n+1$ disjoint parts.

Is there a method how can one do that?

Consider the familiar sequence of Fibonacci numbers: $ F_0=0, F_1=1, F_n=F_{n-1}+F_{n-2}$ .

Although it is rather easy to furnish an algebraic verification of the below identity, I wish to see a different approach. Hence,

QUESTION.Is there a combinatorial or more conceptual reason for this “pretty” identity? $ $ F_nF_{n-1}F_{n-2}=\frac{F_n^3-F_{n-1}^3-F_{n-2}^3}3.$ $

**Caveat.** *I’m open to as many alternative replies, of course.*

**Remark.** The motivation comes as follows. Define $ F_n!=F_1\cdots F_n$ and $ F_0!=1$ . Further, $ \binom{n}k_F:=\frac{F_n!}{F_k!\cdot F_{n-k}!}$ . Then, I was studying these coefficients and was lead to $ $ \binom{n}3_F=\frac{F_n^3-F_{n-1}^3-F_{n-2}^3}{3!}.$ $

Based on online info, it seems that most of these numbers have many solutions. Are there any that have only 1 known solution or only a few solutions?

So, for example, say there’s a cube with side length L, and you drill a cylinder with a diameter larger than L through the cube, what volume remains?

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