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

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 obtained. From what is it derived, from $a^2-b^2$? I recognize that the outcome is the distinction of cubes formula, however searching for it on the internet i only get exercises wbelow the formula is currently provided. Can someone please suggest me in the ideal 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 ehigh quality is the typical trick of adding/subtracting things till you get somepoint you desire. It is normally not terribly insightful but is nonethemuch less conventional. In general, one can variable $a^n-b^n$ as$$ a^n-b^n = (a-b)(a^n-1 + a^n-2b^1 + a^n-3b^2 + cdots + a^2 b^n-3 + a b^n-2 + b^n-1).$$One way of seeing this even more general outcome is as follows: Consider the polynomial equation $x^3-b^3$ (where I have simply supplied $x$ instead of $a$!). Clat an early stage $x=b$ is a root of this equation, so we need to have the ability to aspect out an $(x-b)$ term from the polynomial. Polynomial long division then provides the preferred outcome. This deserve to be generalised to greater $n$.

You are watching: Factor a 3 b 3

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answered Sep 4 "13 at 20:36

Tyler HoldenTyler Holden

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Being able to easily derive the factorization for a distinction of cubes (and also the majority of other things) prevented me from having actually to memorize many kind of things in school. I cannot help however share this with you as it was something I must have done a hundred times in a pinch. I will certainly answer your question by delivering a appropriate full derivation of the conventional factorization for a difference of cubes. Close inspection of the outcome will answer your question I am sure. Begin by broadening $(a-b)^3$ in any manner you choose.

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

We will currently fix this equality for $a^3-b^3$ and also manipulate the various other side of the ehigh quality into the typical formula.

eginalign* (a-b)^3 &= a^3-3a^2b+3ab^2-b^3 \Rightarrowhead 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).endalign*

This is one handy derivation. The amount of cubes deserve to be acquired in a comparable 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 etop quality is obtained by initially grouping terms in the middle expression and then factoring the grouped terms:

$$eginaligna^3-a^2b+a^2b-ab^2+ab^2-b^3&=(a^3-a^2b)+(a^2b-ab^2)+(ab^2-b^3)cr&=a^2(a-b)+ab(a-b)+b^2(a-b)crendalign$$

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 2 $0$"s in between $a^3$ and $-b^3$ (the expressions $-a^2b+a^2b$ and also $-ab^2+ab^2$ are both obviously equal to $0$), but when you perform, lo and behold, the regrouping and factoring work-related their magic.

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answered Sep 4 "13 at 22:16

Barry CipraBarry Cipra

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I am not certain I gain your question. The last formula is just acquired factoring a and b. For example, the initially part

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

So when you acquire the last formula you can take $(a-b)$ and obtain

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

and also to acquire the second formula you sindicate include and also subtract the very same amounts, in specific $-a^2b+a^2b$ and $ab^2-ab^2$.

Does this answer your question?

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answered Sep 4 "13 at 20:39

UmbertoUmberto

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