



 
Hi Bill. A difference of cubes is of course a perfect cube minus a perfect cube. I assume your student already has the formula
We can prove this using polynomial division. First, we look at the roots of a^{3}  b^{3} and immediately we can see that if a = b, then a^{3}  b^{3} = 0, so (ab) is a factor. Next, rewrite it: a^{3}  b^{3} = a^{3} + 0a^{2}b + 0ab^{2}  b^{3} and use synthetic (or long) division to find the other factor. I'll use long division: You could do something very similar for a sum of cubes.
However, there is a geometric way of seeing this too, at least where a and b are positive and a > b: For example, say a= 4 and b = 2. Here is 4^{3}  2^{3}: and here is 3^{3}  2^{3} : Let's take apart 3^{3}  2^{3} first. The red object is broken into the other colours: So the red object is composed of the orange piece which is 3^{2} plus the yellow piece which is 3x2 and the gold piece which is 2^{2}. So the original 3^{3}  1^{3} equals the components 3^{2} + 3x2 + 2^{2}. Now since (32) = 1, we can multiply by this and not change anything, so we've shown
But what about trying something harder, like 5^{3}  2^{3}? We have the same breakdown, but there are 3 copies of everything! This makes sense because (52) = 3. So we've shown that
just as the formula predicted and our long division proved algebraically. If you look carefully at the last diagram, you can consider it as a^{3}  b^{3} and show that the number of copies is always (ab), thus proving the general case. I'd encourage your student to explore in this way the sum of cubes and see if she can come up with a similar algebraic and geometric solution. Cheers,  


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