I recently got interested in jonathanlewisforcongress.comematics 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?

algebra-precalculus

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edited Sep 7 “13 at 10:58

bryanph

asked Sep 4 “13 at 20:30

bryanphbryanph

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## 4 Answers 4

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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 $n$.

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answered Sep 4 “13 at 20:36

Tyler HoldenTyler Holden

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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.

egin{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

ight) \ &=(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.

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edited Sep 4 “13 at 21:56

answered Sep 4 “13 at 21:37

J. W. PerryJ. W. Perry

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

$$egin{align}a^3-a^2b+a^2b-ab^2+ab^2-b^3&=(a^3-a^2b)+(a^2b-ab^2)+(ab^2-b^3)cr&=a^2(a-b)+ab(a-b)+b^2(a-b)crend{align}$$

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.

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answered Sep 4 “13 at 22:16

Barry CipraBarry Cipra

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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 $ab^2-ab^2$.

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answered Sep 4 “13 at 20:39

UmbertoUmberto

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