Associated Topics || Dr. Math Home || Search Dr. Math

### Cubes in a Big Cube

```
Date: 03/11/2002 at 09:45:48
From: Yannik
Subject: Cubes in a big cube

Dr. Math,

I know the formula to find out the number of squares in an n*n square,
which is n(n+1)(2n+1)/6. Is there a formula for the number of cubes in
an n*n*n cube?

Thanks.
```

```
Date: 03/11/2002 at 12:17:37
From: Doctor Peterson
Subject: Re: Cubes in a big cube

Hi, Yannik.

You probably know that the formula you gave is the formula for the sum
of consecutive squares:

1 + 4 + 9 + ... + n^2 = n(n+1)(2n+1)/6

This is the number of squares in an n*n square, because it contains
1 n*n square, 2^2 (n-1)*(n-1) squares, and so on down to n^2 1*1
squares.

The same is true for a cube; you want to find

1 + 8 + 27 + ... + n^3

You can find this by searching our archives for "sum cubes":

Triangular Numbers in a Proof
http://mathforum.org/dr.math/problems/samflan.4.8.97.html

It is

1 + 8 + 27 + ... + n^3 = (1 + 2 + 3 + ... + n)^2

= [n(n+1)/2]^2

For example, a 3*3 cube has 1 3*3, 8 2*2, and 27 1*1 cubes, making a
total of 36, which is the same as [3*4/2]^2 = 6^2.

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Geometry
High School Polyhedra
High School Sequences, Series

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search