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Difference of Two Cubes


Date: 11/18/96 at 20:48:48
From: James Fletcher
Subject: Difference of two cubes

Dear Dr. Math

I am wondering if there is an algebraic method to solve the following 
problem:

The difference of two cubes is 56,765. What two positive integers 
satisfy this condition?

Thank you very much,

James Fletcher


Date: 11/21/96 at 19:18:37
From: Doctor Pete
Subject: Re: Difference of two cubes

I'm not sure what you mean by "algebraic," but you can use some 
algebra to obtain some additional conditions on the solution, if one 
exists.  In particular, note that the difference of two cubes can be 
factored:

     a^3 - b^3 = (a-b)(a^2 + ab + b^2)

So let's assume that a > b.  If a^3 - b^3 = 56765, then if an integer
solution exists, a-b and a^2 + ab + b^2 must be factors of 56765.  
Indeed, we see that 56765 = 5(11353), and this latter factor is prime 
(a bit of arithmetic verifies this).  Therefore, we have:

a-b = 1, a^2 + ab + b^2 = 56765, or a-b = 5, a^2 + ab + b^2 = 11353

  (note that a-b < a^2 + ab + b^2, 
   because a^2 + ab + b^2 = (a-b)^2 + 3ab.)
 
If we assume the first case, then a = b+1, so

     (b+1)^3 - b^3 = 3b^2 + 3b + 1 = 56765

or 3b^2 + 3b = 56764.  But the lefthand side has 3 as a factor, 
whereas the righthand side is not divisible by 3 (the sum of the 
digits is 38).  Hence this leads to an impossibility, so this case is 
not true.  

In the second case, we have a = b+5, and thus

     (b+5)^3 - b^3 = 15b^2 + 75b + 125 = 56765

or b^2 + 5b - 3776 = 0.  At this point, we may use the quadratic 
equation to solve for b.  We obtain:

     b = (-5 + Sqrt[25+4(3776)])/2 = (-5+123)/2 = 59,

(I have omitted the negative root, since the question asks for only 
positive integer solutions).  Thus b = 59, a = 59+5 = 64, and indeed, 
we can verify that

     64^3 - 59^3 = 262144 - 205379 = 56765.

The crucial point here is to notice that the difference of cubes can 
be factored, and you can try to match up factorizations of the given 
integer with these.  It doesn't always work, and sometimes, it works 
in several cases (unlike here, where it works in exactly one case).  
So it's important to try them out.  Furthermore, notice that the 
factorization essentially gives us two equations in two unknowns, and 
that the difference of cubes here allows the degree of these equations 
to be small enough to work with.  If one were given something like the 
difference of fifth powers, then we would not necessarily be able to 
complete the last step, which relies on the fact that we get a 
quadratic equation which is easily solved.

Of course, we get a solution, but it doesn't strike me as being a 
method that guarantees an answer; this is why I hesitate to call it a 
*definitive* solution.  All I did was use some additional facts about 
differences of cubes to obtain extra conditions on a and b so that 
they could be determined - as I mentioned above, this doesn't always 
work.

Hope this helps,

-Doctor Pete,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Basic Algebra
High School Polynomials

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