Help with Problem Investigating Structures Made of Unit Rods

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The discussion focuses on investigating the structure of cubes made from unit rods, specifically calculating the number of joints in various sized cubes. The user seeks a formula to determine the number of joints for cubes like the 3 by 3 by 3 and larger sizes, such as a 55 by 55 by 55 cube. They provide initial data for 1 by 1 by 1 and 2 by 2 by 2 cubes, noting the consistent presence of eight three-joints across all sizes. The challenge lies in deriving a formula that accurately reflects the number of four, five, and six joints as the cube size increases. Assistance is requested to clarify these calculations and develop a comprehensive formula.
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Hey. I’m having trouble with this work I’ve been set about structures. I understand some of it, which I’ll explain later, but I really need your help with it!

My task is to investigate different sized cubes, made up of single unit rods and justify formulae for the number of rods and joints in the cubes. For example, 2 by 2 by 2 cube, 3 by 3 by 3 cube etc etc etc…

I desperately need help on getting a formula to work this out. I know already, as my teacher gave us these results, that a 1 by 1 by 1 cube has the following:

8 Three-Joints ( as do all cubes)
0 Four Joints
0 Five Joints
0 Six Joints

and a 2 by 2 by 2 has:

8 three joints
12 four joints
2 five joints
1 six joints

I really need a formula in order to calculate how many joints a 3 by 3 by 3 has, and with that formula to use it on let's say a 55 by 55 by 55 cube.

All help would be appreciated!
 
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Sorry for being confusing, but here's the task:

Sorry, I was vague.
---
Rigid Structures are constructed using unit rods.

An example is shown below:

* Pic opf a 2 by 2 by 2 cube*

The individual rods in the structures are held together using different types of joints. (Just think of making cubes using matchsticks and playdough*.

http://img61.imageshack.us/img61/6166/joints3lz.jpg

A three joint would be on the vertices of each cube no matter what the size and so therefore there'll be 8 three-joints.
 
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