Factoring $a^3-b^3$

Question

I recently got interested in mathematics after having slacked through it in highschool. Therefore I picked up the book "Algebra" by I.M. Gelfand and A.Shen

At problem 113, the reader is asked to factor $a^3-b^3.$

The given solution is: $$a^3-b^3 = a^3 - a^2b + a^2b -ab^2 + ab^2 -b^3 = a^2(a-b) + ab(a-b) + b^2(a-b) = (a-b)(a^2+ab+b^2)$$

I was wondering how the second equality is derived. From what is it derived, from $a^2-b^2$? I know that the result is the difference of cubes formula, however searching for it on the internet i only get exercises where the formula is already given. Can someone please point me in the right direction?

There are already three answers, so I'll just mention that you might find it interesting to google factoring by grouping. I think it's an interesting method because as a side product it gives students practice in working with binomial common factors. — Dave L. Renfro, Sep 04 '13 at 21:49
Oops, what I meant to suggest looking at is factoring trinomials by the AC method. This is the method I was thinking of that gives, as a side benefit, students practice in working with common factors that are binomials. — Dave L. Renfro, Sep 04 '13 at 22:00

score 10 · Accepted Answer · answered Sep 04 '13 at 20:36

The second equality is the typical trick of adding/subtracting things until you get something you want. It is usually not terribly insightful but is nonetheless standard. In general, one can factor $a^n-b^n$ as $$ a^n-b^n = (a-b)(a^{n-1} + a^{n-2}b^1 + a^{n-3}b^2 + \cdots + a^2 b^{n-3} + a b^{n-2} + b^{n-1}).$$ One way of seeing this more general result is as follows: Consider the polynomial equation $x^3-b^3$ (where I have just used $x$ instead of $a$!). Clearly $x=b$ is a root of this equation, so we should be able to factor out an $(x-b)$ term from the polynomial. Polynomial long division then gives the desired result. This can be generalized to higher $n$.

J. W. Perry · Answer 2 · 2013-09-04T21:56:40.840

Being able to quickly derive the factorization for a difference of cubes (and most other things) prevented me from having to memorize many things in school. I cannot help but share this with you as it was something I must have done a hundred times in a pinch. I will answer your question by delivering a proper full derivation of the standard factorization for a difference of cubes. Close inspection of the result will answer your question I am sure. Begin by expanding $(a-b)^3$ in any manner you choose.

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

We will now solve this equality for $a^3-b^3$ and manipulate the other side of the equality into the standard formula.

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

This is one handy derivation. The sum of cubes can be derived in a similar manner.

score 1 · Answer 3 · answered Sep 04 '13 at 22:16

The second equality is obtained by first grouping terms in the middle expression and then factoring the grouped terms:

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

If the OP is wondering where the middle expression came from in the first place, it's kind of a deus ex machina: All you're doing is sticking two $0$'s between $a^3$ and $-b^3$ (the expressions $-a^2b+a^2b$ and $-ab^2+ab^2$ are both obviously equal to $0$), but when you do, lo and behold, the regrouping and factoring work their magic.

score 0 · Answer 4 · answered Sep 04 '13 at 20:39

I am not sure I get your question. The last formula is just obtained factoring a and b. For example, the first part

$a^2(a-b) = a^3 -a^2b$

So once you get the last formula you can take $(a-b)$ and obtain

$(a-b)(a^2+ab+b^2)$

and to obtain the second formula you simply add and subtract the same quantities, in particular $-a^2b+a^2b$ and $ab^2-ab^2$.

Does this answer your question?

Factoring $a^3-b^3$

4 Answers4