A Different 'View' of Sums of Cubes? An Algebraic "Proof Without Words!"

A well-known and intriguing formula usually proved by Mathematical Induction states that
1³ + 2³ + 3³ + ... +n³ = (1+2+3+...+n)² .

In words:
The sum of the cubes of the first n positive integers equals the square of the sum of the first n positive integers (or the square of the nth triangular number).

Students as early as middle school can investigate numerical patterns of sums of powers of positive integers and can be led to such discoveries. However, in this post we will look at a different kind of "proof." Proofs without words can be fascinating, challenging and can develop a student's spatial reasoning. Just as there have been many visual proofs of the Pythagorean Theorem (dissection type), mathematicians have sought visual arguments for many other numerical patterns and algebraic formulas. The Greeks of antiquity developed many classical arguments of this type, necessitated perhaps by not having our symbolic algebra available.

You will surely find other examples of this on the web (e.g., "Cut-the-Knot") but I thought it might be nice to bring it down to a middle school or Algebra 1 level by having students play with some particular cases of this general formula. I have always been intrigued by this topic, ever since I saw several visual proofs of the Pythagorean Theorem. Later on I was introduced to the genius of Sidney Kung and Roger B. Nelson (Google them!). Prof. Kung's extraordinary visual proofs were (and may still be) a staple of Mathematics Magazine, an MAA publication. You may also recall I have published a couple of other such proofs, one of which came from a student of mine. Look here.

Part I
Let's try to demonstrate that 1³ + 2³ = (1 + 2)²

Before displaying the visual we will begin with an arithmetic-algebraic approach:

Think of (1+2)(1+2) as a special case of the form (a+b)(a+b):
Thus, (1+2)(1+2) = (1⋅1) + (1⋅2) + (2⋅1) + (2⋅2)
Now for some creativity. Since cubes involve a product of THREE factors, we can introduce an extra factor of "1" in each term:
(1+2)(1+2) = (1⋅1⋅1) + (1⋅1⋅2 )+ (1⋅2⋅1) + (1⋅2⋅2).

Even without a visual, we can see the first term on the right is 1³!!
It will take some work to show that the sum of the other three terms is 2³. Ok, with this background, here is a
PROOF WITHOUT WORDS
















Do you think your students will "see" the proof?? My crude attempt at a graphic leaves a lot to be desired! It may be helpful to have manipulatives such as algebra tiles available or have students physically build these models. I would encourage that strongly!

So we are proving a numerical formula using a sum of volumes. You might say we turned squares into cubes!!

Do you think this investigation is through? Of course not -- I did all the work for you. Now here is the real test:

Part II

Show that 1³ + 2³ + 3³ = (1 + 2 + 3)²
using a "Proof Without Words."

Ok, I'll give you a little hint although you don't need to use this:
Rewrite (1 + 2 + 3)²
as ((1 + 2 )+ 3)<sup>2

Have fun! Just think, if we have a sum of 4th powers, we might need hypercubes!</sup>