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Every resource that I've read proves the formula $$ a^3 + b^3 = (a+b)(a^2-ab+b^2) \tag1$$ by just multiplying $(a+b)$ and $(a^2 - ab + b^2)$.

But how did people come up with that formula? Did they think like, "Oh, let's just multiply these polynomials, I don't know why, let's just do it." I don't think that people just pointed a finger at the sky and came up with that formula.

So, how to prove $(1)$?

Blue
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Rodion Iskhakov
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    What do you mean? It's clear that $a^3+b^3$ vanishes if $a+b=0$ so, even without general theory, you might expect that $a+b$ divides $a^3+b^3$. Now it's just a matter of finding the coefficients of the quadratic term. – lulu Sep 05 '22 at 19:27
  • Well, people thought about the reverse problem (finding the sum of a geometric progression) since ancient times. – PM 2Ring Sep 05 '22 at 19:32
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    This follows easily from the well known formula for the finite geometric series $1 + 2 + s + x^2 + \cdots + x^n = (x^{n+1} -1)/(x-1)$. That one you could guess from numerical evidence with $=2$ and some other values. – Ethan Bolker Sep 05 '22 at 19:34
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    People will have known about $a^2-b^2=(a+b)(a-b)$ and probably sought to consider $a^3-b^3$ or perhaps $a^3+b^3$, as in your case. Mathematicians are explorers too. – pshmath0 Sep 05 '22 at 19:50
  • @Radion Ishakov : Divide $x^3+1 $by $x+1$ by long division to obtain the coefficients(1,−1,1) .. so make a quadratic factor with (a,b) – – Narasimham Sep 05 '22 at 19:58
  • Not exactly your question but you might find this interesting - it describes the progress of solving the general cubic equation in the 1500s - https://www.quantamagazine.org/the-scandalous-history-of-the-cubic-formula-20220630/ – Mike O'Connor Sep 06 '22 at 01:30
  • Example $10^3+(-1)^3 = 1000 - 1 = 999 = 9\cdot 111 = (10+(-1))(10^2+10+1)=(10+(-1))(10^2-10(-1)+(-1)^2).$ The general result is almost immediate. – Somos Sep 06 '22 at 11:49

4 Answers4

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$«$Who first factored the expression $a^3+b^3$ and what was the method used?$»$ - I don't know the exact answer to this question, but the similar question I'm trying to answer is:

$«$How can we factor $a^3+b^3$ using the most basic algebraic techniques?$»$

It seems to me that, this formula basically comes from the Binomial theorem:

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

But, without the binomial theorem you can obtain this result as follows:

$$\begin{aligned}(a+b)^3&=(a+b) \left((a+b)\times (a+b)\right)\\ &=(a+b)(a^2+2ab+b^2)\\ &=a^3+3a^2b+3ab^2+b^3\end{aligned}$$

Then observe that,

$$\begin{aligned}a^3+3a^2b+3ab^2+b^3&=(a^3+b^3)-3ab(a+b)\end{aligned}$$

This leads,

$$ \begin{aligned}a^3+b^3&=(a+b)^3-3ab(a+b)\\ &=(a+b)\left((a+b)^2-3ab\right)\\ &=(a+b)(a^2-ab+b^2).\end{aligned} $$

nonstudent
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As noticed by $a=-b$ we obtain $a^3+b^3=0$ therefore we can guess by homogeneity

$$a^3+b^3=(a+b)(Xa^2+Yab+Zb^2)$$

with $X$, $Y$ and $Z$ unknown, to obtain

$$(a+b)(Xa^2+Yab+Zb^2)=Xa^3+Ya^2b+Zab^2+Xa^2b+Yab^2+Zb^3=$$

$$=Xa^3+(Y+X)a^2b+(Z+Y)ab^2+Zb^3$$

which requires

  • $X=1$
  • $Z=1$
  • $Y+X=0 \implies Y=-1$
  • $Z+Y=0 \implies Y=-1$
user
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    Much more complicated polynomials can be factored with this method. It will be good for the OP to know this. – nonstudent Sep 05 '22 at 19:59
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Alternative approach:
Stealing the insight from the comment of Ethan Bolker.

Mathematicians discovered that

$$(1 + x + x^2 + \cdots + x^n) \times (1 - x) = 1 - x^{n+1}. \tag1 $$

For $0 \neq a,b$,
$a^3 + b^3$ can be rewritten as

$$a^3 \times \left[1 - \left(\frac{-b}{a}\right)^3\right]. \tag2 $$

Setting $~\displaystyle x = \left(\frac{-b}{a}\right)~$ (1) and (2) collectively imply that

$$a^3 + b^3 = a^3 \times \left[1 - \frac{-b}{a}\right] \times \left[ ~1 + \frac{-b}{a} + \left(\frac{-b}{a}\right)^2 ~\right]$$

$$ = a^3 \times \left[\frac{a + b}{a}\right] \times \left[\frac{a^2 - ab + b^2}{a^2}\right] = (a+b) \times (a^2 - ab + b^2).$$

user2661923
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  • I would say that they discovered $a^3 + b^3 = (a+b)(a^2-ab+b^2)$ before than $(1 + x + x^2 + \cdots + x^n) \times (1 - x) = 1 - x^{n+1}$. Anyway this is very informative. – user Sep 05 '22 at 20:45
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In this answer I wanted to suggest a method where I get results by applying substitutions (based on the idea that it is the most basic algebraic technique).

Let, $a=m+n$ and $b=m-n$ then:

$$ \begin{aligned}a^3+b^3=m^3+3m^2n+3mn^2+n^3+m^3-3m^2n+3mn^2-n^3=2m^3+6mn^2=2m(m^2+3n^2)\end{aligned} $$

and note that,

$$ \begin{cases} m+n=a\\m-n=b\end{cases}\implies \begin{cases}m=\frac {a+b}{2}\\n=\frac {a-b}{2}\end{cases} $$

Putting $m=\frac {a+b}{2}$ and $n=\frac {a-b}{2}$, we have:

$$ \begin{aligned}a^3+b^3&=2m(m^2+3n^2)\\ &=2\times \frac {a+b}{2}\left(\frac {(a+b)^2}{4}+3\times \frac {(a-b)^2}{4}\right)\\ &=(a+b)\left(\frac {(a+b)^2+3(a-b)^2}{4}\right)\end{aligned} $$

Now, let's open the parentheses and do some basic simplifications:

$$ \begin{aligned} (a+b)^2+3(a-b)^2&=a^2+2ab+b^2+3(a^2-2ab+b^2)\\ &=a^2+2ab+b^2+3a^2-6ab+3b^2\\ &=4a^2-4ab+4b^2\\ &=4(a^2-ab+b^2)\end{aligned} $$

Thus, we get the following result:

$$ \begin{aligned}a^3+b^3&=(a+b)\times \frac {4(a^2-ab+b^2)}{4}\\ &=(a+b)(a^2-ab+b^2).\end{aligned} $$

nonstudent
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