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# A Cubic Identity

Alignments to Content Standards: A-SSE.B.3.a A-SSE.A.2

Let $a$ and $b$ be real numbers with $a>b>0$ and $\frac{a^3-b^3}{(a-b)^3}=\frac{73}{3}$. What is $\frac{b}{a}$?

## IM Commentary

This task presents a challenging exercise in both algebraic manipulations and seeing structure in algebraic expressions (MP 7). In the solutions provided, students have to identify that $(a-b)$ is a factor of $(a^3-b^3)$ in order to simplify the expressions appearing to quadratics. At that point, either factoring or the quadratic formula can be employed to compute $\frac{b}{a}$.

This task was adapted from problem #17 on the 2012 American Mathematics Competition (AMC) 10A Test, which asked students for the value $a-b$ given that $a$ and $b$ were relatively prime integers. (In the end, this forces $a=10$ and $b=7$, so $a-b=3$.) For the 2012 AMC 10A, which was taken by 73,703 students, the multiple choice answers for the problem had the following distribution:

 Choice Answer Percentage of Answers (A) 1 6.15 (B) 2 5.11 (C) 3 25.76 (D) 4 5.08 (E) 5 2.44 Omit -- 55.42

Of the 73,703 students: 36206, or 49%, were in 10th grade;25,498 or 35%, were in 9th grade; and the remainder were below than 9th grade.

## Solutions

Solution: 1 Algebra, then Factoring

Vital to solving this problem is the fact that the numerator is a difference of cubes, which suggests using the identity $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$. Using this identity, and that $a\neq b$, we can simplify the expression

\begin{align} \frac{a^3 - b^3}{(a - b)^3} &= \frac{(a-b)(a^2 + ab + b^2)}{(a-b)(a^2 -2ab + b^2) } \\ &= \frac{ a^2 + ab + b^2 }{ a^2 - 2ab + b^2 }. \end{align}

The equation $\frac{a^3-b^3}{(a-b)^3} = \frac{73}{3}$ can be written, using the identity from the previous paragraph, as as

\begin{align} 3(a^2 + ab +b^2) = 73(a^2 - 2ab + b^2). \end{align}
Gathering all terms on the same side, this is the same as $70a^2-149ab+70b^2 = 0$. Fortunately, this quadratic can be factored
\begin{align} 70a^2-149ab+70b^2 = (10a - 7b)(7a -10b). \end{align}

Because $a \gt b \gt 0$, we are looking for a solution to $7a - 10b = 0$ rather than $10a - 7b = 0$, which gives $7a=10b$. Re-writing, we find $$\frac{b}{a}=\frac{7}{10}=0.7.$$

Solution: 2. Algebra, then Quadratic Formula

This solution is identical to the first until the quadratic relation between $a$ and $b$ is reached, at which point we instead use here the quadratic formula.

Vital to solving this problem is the fact that the numerator is a difference of cubes, which suggests using the identity $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$.

Using this identity, and that $a\neq b$, we can simplify the expression

\begin{align} \frac{a^3 - b^3}{(a - b)^3} &= \frac{(a-b)(a^2 + ab + b^2)}{(a-b)(a^2 -2ab + b^2) } \\ &= \frac{ a^2 + ab + b^2 }{ a^2 - 2ab + b^2 }. \end{align}

The equation $\frac{a^3-b^3}{(a-b)^3} = \frac{73}{3}$ can be written, using the identity from the previous paragraph, as as

\begin{align} 3(a^2 + ab +b^2) = 73(a^2 - 2ab + b^2). \end{align}

Gathering all terms on the same side, this is the same as $70a^2-149ab+70b^2 = 0$. Dividing both sides by $a^2$, we are left with the quadratic equation $$70\left(\frac{b}{a}\right)^2-149\left(\frac{b}{a}\right)+70=0.$$ Using the quadratic formula gives $$\frac{b}{a}=\frac{149\pm\sqrt{149^2-4\cdot 70^2}}{140}=\frac{149\pm 51}{40}=\frac{10}{7}\text{ or }\frac{7}{10}.$$ Since $a \gt b \gt 0$, $\frac{7}{10}$ must be the solution we are looking for.

Alternatively, we could apply the quadratic formula directly to the $(70)a^2-(149b)a+(70b^2)=0$ to get $$a=\frac{149b\pm \sqrt{(149b)^2-4\cdot 70\cdot (70b^2)}}{140}=\frac{(149\pm 51)b}{140},$$ giving to the same solution.

#### Gwyneth says:

almost 2 years

Kassymov makes an excellent point about the "trick" to this problem being the identity that allows us to rewrite a^3-b^3. This approach is fine if the goal is to develop or utilize such identities, but is unlikely to engage many students. Realistically, if one needed to solve such a problem for a contextual reason one would already be sitting in front of a computer. So let’s think about how we could take advantage of that fact.

We know that the ratio b/a should be constant, regardless of the value of a. So let’s just assume a is 1 and then figure out what b could be. Plugging in 1 for a and then moving all terms to one side of the equation gives us: 0=73(1-b)^3 – 3 + 3b^3. Graphing this as a function in desmos, I found that the zeroes were located at b=0.7, 1, and 1.429. Given that a is greater than b and a=1, I know that b=0.7, so the ratio b/a=0.7.

#### Cam says:

almost 2 years

Fair point, though if we're sitting in front of a computer, couldn't we even more directly ask Wolfram alpha?

http://www.wolframalpha.com/input/?i=solve+for+b+in++%28a%5E3-b%5E3%29%2F%28a-b%29%5E3%3D73%2F3

It seems to me your objection is different from kassymov's (though equally valid), in that the objection is the presence of a test-taking trick to get the answer (the knowledge that the answer exists helps you find it).

I'd certainly concede the point that this trick exists, though I'm not sure that this diminishes the quality of the task terribly. The content behind many algebraic tasks could easily be circumvented by the presence of computers, and so the question is whether there is merit in being able to do the question without them. Your comment seems to suggest that you believe the answer to be no (in this case -- not suggesting that you are dismissing algebra as a whole), but I think there's room for debate as to how "trick"-y this problem really is. The fact that differences of squares factor nicely is a "trick," but I think one that a vast majority of us agrees is so pervasive as to merit its inclusion in just about everyone's algebra toolbox. That differences of cubes (or any higher power) do as well does not seem to me to be far-fetched, though I could certainly envision reasonable people disagreeing on this front.

#### Gwyneth says:

almost 2 years

Good points. I suppose my bigger point is that there are multiple solution paths to such problems. Acknowledging this allows for classroom differentiation and discussion of when different methods are most appropriate. When does it make sense to find a solution with algebraic manipulation? When does one utilize less precise methods like graphing? How do we interpret an answer generated by technology? and so on...

#### Cam says:

almost 2 years

Completely agreed, though I'd maybe argue that your bigger point is so fundamental to mathematics and the teaching of mathematics that it's hard to play out all of its implications in every task. For this task in particular, I think the graphical approach is actually a little on the awkward side. If the prompt were slightly more open-ended ("What can you say about $b/a$"?), then plugging in a=1 and learning b=.7 tells you very little about the value of $b/a$ in general. (It's only the question giving away that $b/a$ has a constant numerical value that lets the trick work with the current prompt). Perhaps this is evidence that the prompt should be changed, which is certainly up for debate, though I lean toward leaving it as is, and letting the task stand as an instance where seeing the structure of the algebraic expression leads the path to the solution, even if it is not the unique such path.

#### kassymov says:

over 3 years

What is the significance that a and b are integers? Seems like only the a > b condition is used.

A more philosophical question. How valuable is it to consider examples of this kind? I can see how AMC participants would think that an "ugly" expression is likely to be factor nicely. However, 73 and 3 had to be carefully selected. Even for this type of problem, the method works in relatively few cases. Could it be that using examples like this for illustration would reinforce the idea that mathematics is about tricks coming from nowhere and applied randomly. (I can see how a math Olympiad participant can use a bag of tricks to try them one by one until the problem is solved, but that's a special group of students).

#### Cam says:

over 3 years

Thanks very much for these comments. A mix-up on my part had the commentary and problem heading of the original AMC problem, while the solutions were those of a re-write on our part. So the integer condition is certainly not necessary.

I think the new commentary may address your (very valid) philosophical question. At the end, this is a non-trivial exercise in algebraic thinking and manipulation. But only in that one solution is it necessary that 7 and 3 were carefully selected. I've written a second solution using the quadratic formula which would work for any other pair of values as well, which I think at least somewhat diminishes the extent to which this is just a "bag of tricks" problem. (Of course, bags of tricks are also fairly valuable things to acquire along the way...)

In any case, further feedback very welcome. Thanks again.

#### Boaz Munro says:

almost 4 years

I may be missing something, but it looks like the correct answer (.7) was not listed as an option on the AMC test!

#### Cam says:

almost 4 years

Hi boazmunro,

Just missing something silly -- the commentary notes that (as it went through the IM review process) the problem was modified from asking about a-b to asking about a/b. The AMC test asked for a-b, for which the correct answer is (c) 3. As a small amount of the backstory, the original problem required students to reason using the concept of "relatively prime" integers -- the current version much more heavily focuses on the algebraic structure of the expressions involved, so as to better illustrate A-SSE.A.2.