How to motivate $x^n−y^n ≡ (x−y)(x^{n−1}+x^{n−2}y+...+xy^{n−2}+y^{n−1})$, to 13 year olds?

Question

You can safely presuppose that 13 year old (y.o.) students learned the Difference of Squares and of Cubes identities, before tackling this Difference of Powers Identity ($x^n – y^n$).

The glut of duplicates proves that students need better motivation. How can teachers demystify the 2 key steps below?

Too many students can’t even forebode the 1st step, which is the Sum of a Geometric Sequence: $\color{peru}{r^n -1 ≡ (r – 1)(r^{n-1}+r^{n-2}+...+r^2+r+1)}$
Step 2 is even more Delphic! The Difference of Powers Identity has no division! Then how can students foreknow to substitute $r = x/y$, into the above $\color{darkgoldenrod}{\text{Sum of a Geometric Sequence}}$?

I’m NOT asking how to prove with algebraic expansion. This identity is exercise 1v in Michael Spivak, Calculus (4th edn 2008), page 13.

In what context is this being taught to 13 year olds? In the standard American curriculum, difference of squares and difference of cubes might come up in a student's sophomore high school year (15-16 yo), but I believe it is more commonly taught in the junior year (16-17 yo). In that context, they will likely not have seen series yet, and the identities are often presented with little justification or proof. — Xander Henderson, Oct 15 '23 at 20:09
Personally, I think a 13-year old student who is advanced enough for something like this would not need motivation, unless maybe the reason the student is so advanced is that the student was heavily pushed externally rather than reached this level on their own initiative. — Dave L Renfro, Oct 15 '23 at 20:10
A slightly more advanced instructor might note that $a$ is a root of $x^n - a^n$, hence $x-a$ is a factor. Hence one can use polynomial long division to work out the other factor. Here, the instructor might make reference to some kind of "pattern matching" or (in a more advanced setting) a proof-by-induction style argument. — Xander Henderson, Oct 15 '23 at 20:12
@XanderHenderson I'm not American. This identity is usually taught to 13 year olds in my country. Kindly elaborate your second "note that a is a root of $x^n - a^n$" and "polynomial long division"? — user95017, Oct 15 '23 at 22:38
@SueVanHattum I am? What was first time? I never asked about $x^n - y^n$ here? — user95017, Oct 15 '23 at 22:39
I'd start my search for a way to motivate the formula with the question, Why does a 13-year-old need to know this? It made the difference of cubes easier to remember for me, but I was older than 13 at that point. A really cool way to see it is to start with an $n$-cube of side length $x$ and slice slabs of thickness $x-y$ off each side opposite a fixed base point in succession. Of course, you can't really see once $n$ is greater than $3$. Well — sorry to flip-flop — you could see it if you projected it into ${\bf R}^3$ first, which is possible. I don't think a 13yo needs to know this, tho — user1815, Oct 15 '23 at 22:39
@user95017 "I'm not American." Which is why I asked you to give context. This is all stuff that should be in your question. Explain your country's curriculum. Why is this being taught to 13 year olds? — Xander Henderson, Oct 15 '23 at 22:50
It's really not clear to me what you mean by "motivation." Are you looking to provide students with reasons for why this identity is useful/important? An understandable proof? Connections to geometric/physical intuition? Mnemonic devices? — Justin Hancock, Oct 15 '23 at 23:07
@JustinHancock I write "motivation" because I see it so much on Math SE! But is there a better word? What word capsulizes all these quotations? "If one were developing the theory from scratch, how would one find this solution (other than blind luck)?" I seek "good explanations and motivations for everything (as opposed to just pulling out ready-made solutions like what was done to me when I was learning this exact thing)".. — user95017, Oct 15 '23 at 23:47
@JustinHancock "This trick may seem mysterious. After all, why did 'multiplying by the conjugate' work, and it is always possible to find such tricks?" "What is the motivation for this formula and how can I get there? I have been told about how the 2x2 formula is guessable, but this is no convincing argument for higher dimensions.". — user95017, Oct 15 '23 at 23:47
@JustinHancock See also Motivation for taylor series of e, Motivation for Ramanujan's mysterious π formula, Motivation for definition of linearity, — user95017, Oct 15 '23 at 23:48
Seriously, where in the world is this? This isn't even in the curriculum for a 2-year college math degree where I teach. Very curious. — Daniel R. Collins, Oct 16 '23 at 01:31
Oops. I will delete my comment, and yours, and this, after I expect you've seen it. — Sue VanHattum, Oct 16 '23 at 03:22
I first encountered this with synthetic division (late 1973; taking 9th grade Algebra 1 at age 14-15, but I was simultaneously and individually working through our school's 11th grade Algebra 2 text at the time), when one of the more difficult Algebra 2 problems was to find the quotient when $x^n - a^n$ is divided by $x-a.$ I recall checking the result with $x^4 - a^4 = (x^2 - a^2)(x^2 + a^2) = (x-a)(x+a)(x^2 + a^2),$ and noticing that it followed a pattern like the binomial theorem except the coefficients were about as simple as possible, and it was easy to see why it was true by multiplying. — Dave L Renfro, Oct 16 '23 at 08:33
@DanielR.Collins I now teach in China, but I have taught in many other countries. — user95017, Oct 17 '23 at 00:31
Thanks. Are you saying that this kind of topic is on the curriculum for 13 year olds in all of those many countries? — Daniel R. Collins, Oct 17 '23 at 06:16

score 6 · Answer 1 · answered Oct 16 '23 at 21:25

Start off by observing that:

$99=9 \times 11$

$999=9 \times 111$

$9999=9 \times 1111$

$99999=9 \times 11111$

The left hand side is a power of ten minus one. The second number on the right hand side is a sum of all the powers of ten up to but not including the one on the left. We obviously used base 10 here, but we can pull this trick in any base. Numbers of the form $111...1$ are a sum of powers of the base. We multiply by the base minus one to get as high as we can without carrying, and then add one to trigger the carry and get the next base power. We can write this:

$(b-1)(b^{n-1}+b^{n-2}+\ldots b^2+b^1+b^0)=b^n-1$

We can prove this by just multiplying out the brackets, and observing that all but the first and last terms cancel.

Then point out that the multiplying out brackets proof means this works for any number $b$, not just whole numbers suitable for use as a base. Fractions $b=x/y$ are then an obvious thing to try out, to demonstrate that. Finally, multiply that by $y^n$ to cancel the fractions, and get the desired result.

score 5 · Answer 2 · answered Oct 15 '23 at 22:12

Since you're asking about teaching the general difference-of-$n$th-powers formula, I'll assume the students have already learned the difference-of-squares and difference-of-cubes formulas:

$$\begin{align*} n=2: \quad x^2 - y^2 &= (x - y)(x + y)\\ n=3: \quad x^3 - y^3 &= (x - y)(x^2 + xy + y^2) \end{align*}$$

If they haven't learned these yet, then something is very wrong with either 1) their curriculum progression leading up to this point, or 2) the expectation that they should be able to learn the general difference-of-$n$th-powers formula.

To nudge them towards the general formula, I'd rewrite the above with color-coding and all exponents explicit:

$$\begin{align*} n=2: \quad \color{blue}{x^2} - \color{red}{y^2} &= (\color{blue}{x} - \color{red}{y})(\color{blue}{x^1}\color{red}{y^0} + \color{blue}{x^0}\color{red}{y^1}) \\ n=3: \quad \color{blue}{x^3} - \color{red}{y^3} &= (\color{blue}{x} - \color{red}{y})(\color{blue}{x^2}\color{red}{y^0} + \color{blue}{x^1}\color{red}{y^1} + \color{blue}{x^0}\color{red}{y^2}) \end{align*}$$

Then I'd ask them to describe the pattern to me. (If they can't do this, then I'd use Socratic questioning to get them over the hump.)

Once they've successfully described the pattern, I'd ask them to take their best guess at the case of $n=4.$ We'd multiply the RHS out the long way to verify it.

$$n=4: \quad \color{blue}{x^4} - \color{red}{y^4} = (\color{blue}{x} - \color{red}{y})(\color{blue}{x^3}\color{red}{y^0} + \color{blue}{x^2}\color{red}{y^1} + \color{blue}{x^1}\color{red}{y^2} + \color{blue}{x^0}\color{red}{y^3})$$

And then I'd ask them to write down the formula for arbitrary $n.$

$$\color{blue}{x^n} - \color{red}{y^n} = (\color{blue}{x} - \color{red}{y})(\color{blue}{x^{n-1}}\color{red}{y^0} + \color{blue}{x^{n-2}}\color{red}{y^1} + \cdots + \color{blue}{x^1}\color{red}{y^{n-2}} + \color{blue}{x^0}\color{red}{y^{n-1}})$$

thanks! Yes, you are correct to "assume the students have already learned the difference-of-squares and difference-of-cubes formulas". I just added this presumption to my question. — user95017, Oct 15 '23 at 22:36
I know what Socratic questioning means, in the main. But I don't know how to apply it here here..."(If they can't do this, then I'd use Socratic questioning to get them over the hump.)" Kindly elaborate? Can you write out this Socratic dialogue? — user95017, Oct 17 '23 at 00:28
@user95017 Q: In the second factor on the RHS, what are the exponents of the $x$'s, from left to right?" A: $2,1,0.$ Q: What are the exponents of the $y$'s, from left to right? A: $0,1,2.$ Q: In any given term, what do the exponents add to? $2.$ — Justin Skycak, Oct 17 '23 at 00:58

score 4 · Accepted Answer · answered Oct 16 '23 at 15:29

Here a slightly different proposal.

Ask the students to find $A_1$, $A_2$, ... , $A_n$ such that the formula

$$ x^n - y^n = (x-y)(A_1+A_2+ \dots + A_n) $$

holds.

You can get that $A_1 = x^{n-1}$ such that the product by $x$ gives $x^n$. But doing this will generate an extra factor $B_1= -y \cdot A_1 = -x^{n-1}y$ when multiplied by $-y$. We can however choose $A_2$ such that it cancels out $B_1$: $A_2 = x^{n-2}y$. When multiplied by $x$, $A_2$ cancels out exactly $B_1$, but it generates an extra factor $B_2 = -y \cdot A_2=-x^{n-2}y^2$ when multiplied by $-y$.

This in turns generates $A_3 = x^{n-3}y^2$ and so on.

Eventually we arrive at $A_n = y^{n-1}$ that cancels out $B_{n-1}$ when multiplied by $x$ and generates the required factor $-y^n$ when multiplied by $-y$.

(+1) Nice! Probably when showing this to students it would be best to do this for a couple of specific values of $n,$ such as $n=4$ and $n=7.$ (I suspect using both an even integer for $n$ and an odd integer for $n$ would be better than using two evens or using two odds.) — Dave L Renfro, Oct 16 '23 at 17:16

score 2 · Answer 4 · answered Oct 17 '23 at 21:33

2

Would long division help? I had something like this in high school

$$ \require{enclose} x - y \enclose{longdiv} {x^2 \qquad- y^2} $$

We then figured out the answer x + y. You can increase the powers of the dividend to 3, then 4, and eventually n.

answered Oct 17 '23 at 21:33

Daniel

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How to motivate $x^n−y^n ≡ (x−y)(x^{n−1}+x^{n−2}y+...+xy^{n−2}+y^{n−1})$, to 13 year olds?

4 Answers4