The minus difference between cubes of two natural numbers is 208

1. Jul 9, 2010

Ryker

1. The problem statement, all variables and given/known data
The difference between cubes of two natural numbers is 208. Which are those two numbers?

2. Relevant equations
-

3. The attempt at a solution
Here is how I set it up.

x$$\widehat{}^{}3$$ - (x-y)$$\hat{}3 = 208$$, which leads to ...
x$$\hat{}2(3y)$$ + x(-3y$$\hat{}2)$$ + (y$$\hat{}3 - 208)$$ = 0

I figured, since this is now a quadratic equation and I'm looking for a solution where the curve only touches the x-axis, D = b^2 - 4ac would have to be zero. But when I put that in, it just doesn't budge. I know y = 4 and x = 6, but for the love of God, I can't get to that solution. I'm really starting to get nervous, since this is supposed to be an easy problem, but I just can't see what I'm doing wrong to not get to the solution.

Any help would be greatly appreciated.

edit: After charting it out, I see that the determinant isn't zero, so I got that wrong, but I still don't see what I need to do to get y = 4.

Last edited: Jul 9, 2010
2. Jul 9, 2010

HallsofIvy

Staff Emeritus
First, I see no reason why that quadratic should have its vertex at the x-axis nor do I see any reason to call one of the unknowns "x- y". If instead you call them "x" and "y" you have [itex]x^3- y^3= (x- y)(x^+ xy+ y^2). Now look at the ways to factor 208: 2(104)= (4)(52)= (8)(26)= (13)(16). If x- y is equal to the smaller of those factors, is $x^2+ xy+ y^2$ equal to the other?

3. Jul 9, 2010

Ryker

First of all, thanks for the quick reply. I was thinking along the lines you suggested, as well, but dismissed it, because it still seems that, even though you arrive at the correct solution, you need to try out a few options (say, if you don't try 4 x 52 as your first option and go with the 2 x 104, your first attempt is a failure). Is that the only way to do it, though? I mean, is there no elegant solution that would not require any guesswork, but would invariably lead to a single solution without having to try out different options? The reason I'm asking this is because the problem wasn't labelled as being one of the harder ones, and while going with what you suggested is pretty straight-forward, I feel a bit of unease when a solution to a maths problem isn't streamlined so as to know it's right without having to test it.

I don't know, I hope this makes sense and that you get what I intented to say :)

4. Jul 9, 2010

Staff: Mentor

I think so. The only information you are given is that you have two natural numbers whose difference of cubes is 208. I don't see any other approach that to factor x3 - y3 as HallsOfIvy did, and compare this factorization to that of 208.
What guesswork? Since x3 - y3 can be factored in only one way, and 208 can be factored into just a few pairs of numbers (e.g., 1*208, 2*104, etc.), the problem reduces to listing the possible factor pairs and finding the pair that works. This isn't guessing.
I don't get what you're saying here. Whenever you come up with a solution to a math problem, it's always a good idea to check it.

Last edited: Jul 9, 2010
5. Jul 14, 2010

Ryker

Sorry for the late reply, but thanks for your help, I really appreciate it. And I have to say it's somewhat comforting to hear there isn't another way to solve this, because I actually feel less stupid now :)
I guess what I meant was that it isn't like, say, factoring an easy polynom, such as x^2 - 2x + 1 = 0, where you *know* the only solution is (x-1)^2. There isn't much guesswork being done there and, like I said, you know what the solution prior to actually having to check whether (x-1)^2 does in fact equal the original polynom. With the problem I had made a thread about, however, it still seems to me that the checking the solution part and finding it are basically one and the same as opposed to that polynom factoring where they are distinctly separated. Hope this now clears what I meant to say :)

But in any case, my thanks to both of you!