Several years ago, I posed on this ‘blog a problem I learned from Natalie Jackucyn:

For some integers A, B, and n, one term of the expansion of is . What are the values of A, B, and n?

In this post, I reflect for a moment on what I’ve learned from the problem and outline a solution approach before sharing a clever alternative solution one of my students this year leveraged through her CAS-enabled investigation.

**WHAT I LEARNED BEFORE THIS YEAR**

Mostly, I’ve loved this problem for its “reversal” of traditional binomial expansion problems that typically give A, B, and n values and ask for either complete expansions or specific terms of the polynomial. Both of these traditional tasks are easily managed via today’s technology. In Natalie’s variation, neither the answer nor how you would proceed are immediately obvious.

The first great part of the problem is that it doesn’t seem to give enough information. Second, it requires solvers to understand deeply the process of polynomial expansion. Third, unlike traditional formulations, Natalie’s version doesn’t allow students to avoid deep thinking by using technology.

In the comments to my original post, Christopher Olah and a former student, Bryan Spellman, solved the problem via factoring and an Excel document, respectively. Given my algebraic tendencies, I hadn’t considered Bryan’s Excel “search” approach, but one could relatively easily program Excel to provide an exhaustive search. I now think of Bryan’s approach as a coding approach to a reasonably efficient search of the sample space of possible solutions. Most of my students’ solutions over the years essentially approach the problem the same way, but less efficiently, by using one-case-at-a-time expansions via CAS commands until they stumble upon good values for A, B, and n. Understandably, students taking this approach typically become the most frustrated.

Christopher’s approach paralleled my own. The x and y exponents from the expanded term show that n=5+3=8. Expanding a generic then gives a bit more information. From my TI-Nspire CAS,

so there are 56 ways an term appears in this expansion before combining like terms (explained here, if needed). Dividing the original coefficient by 56 gives , the coefficient of .

The values of a and b are integers, so factoring 497,664 shows these coefficients are both co-multiples of 2 and 3, but which ones? In essence, this defines a system of equations. The 3 has an exponent of 5, so it can easily be attributed to a, but the 11 is not a multiple of either 5 or 3, so it must be a combination. Quick experimentation with the exponents leads to , so goes to a and goes to b. This results in and .

**WHAT A STUDENT TAUGHT ME THIS YEAR**

After my student, NB, arrived at , she focused on roots–not factors–for her solution. The exponents of a and b suggested using either a cubed or a fifth root.

The fifth root would extract only the value of a if b had only singleton factors–essentially isolating the a and b values–while the cubed root would extract a combination of a and b factors, leaving only excess a factors inside the radical. Her investigation was simplified by the exact answers from her Nspire CAS software.

From the fifth root output, the irrational term had exponent 1/5, not the expected 3/5, so b must have had at least one prime factor with non-singular multiplicity. But the cubed root played out perfectly. The exponent–2/3–matched expectation, giving a=6, and the coefficient, 24, was the product of a and b, making b=4. Clever.

**EXTENSIONS & CONCLUSION**

Admittedly, NB’s solution would have been complicated if the parameter was composed of something other than singleton prime factors, but it did present a fresh, alternative approach to what was becoming a comfortable problem for me. I’m curious about exploring other arrangements of the parameters of to see how NB’s root-based reasoning could be extended and how it would compare to the factor solutions I used before. I wonder which would be “easier” … whatever “easier” means.

As a ‘blog topic for another day, I’ve learned much by sharing this particular problem with several teachers over the years. In particular, the initial “not enough information” feel of the problem statement actually indicates the presence of some variations that lead to multiple solutions. If you think about it, NB’s root variation of the solution suggests some direct paths to such possible formulations. As intriguing as the possibilities here are, I’ve never assigned such a variation of the problem to my students.

As I finish this post, I’m questioning why I haven’t yet taken advantage of these possibilities. That will change. Until then, perhaps you can find some interesting or alternative approaches to the underlying systems of equations in this problem. Can you create a variation that has multiple solutions? Under what conditions would such a variation exist? How many distinct solutions could a problem like this have?