I recently acquired interested in neurosoup.orgematics after having actually slacked v it in highschool. As such I choose up the book "Algebra" by I.M. Gelfand and A.Shen

At trouble 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 to be wondering exactly how the second equality is derived. Native what is it derived, native $a^2-b^2$? I know that the result is the difference of cubes formula, but searching for it top top the internet i only obtain exercises whereby the formula is already given. Have the right to someone please point me in the ideal direction?

algebra-precalculus
re-superstructure
point out
follow
edited Sep 7 "13 at 10:58
bryanph
asked Sep 4 "13 at 20:30 bryanphbryanph
$\endgroup$
2
include a comment |

9
$\begingroup$
The 2nd equality is the usual trick that adding/subtracting things until you gain something friend want. That is typically not terribly insightful yet is however standard. In general, one can element $a^n-b^n$ 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 method of see this an ext general result is together follows: think about the polynomial equation $x^3-b^3$ (where I have actually just offered $x$ instead of $a$!). Plainly $x=b$ is a root of this equation, so we should have the ability to factor out an $(x-b)$ term native the polynomial. Polynomial long division then provides the preferred result. This deserve to be generalised to greater $n$.

You are watching: Factor a 3 b 3

re-publishing
mention
monitor
answer Sep 4 "13 in ~ 20:36 Tyler HoldenTyler Holden
$\endgroup$
include a comment |
2
$\begingroup$
Being may be to conveniently derive the factorization for a difference of cubes (and most other things) impede me from having to memorize countless things in school. Ns cannot assist but re-publishing this through you together it was something ns must have actually done a hundred times in a pinch. I will answer your inquiry by transferring a suitable full source of the conventional factorization for a difference of cubes. Close inspection of the result will price your concern I to be sure. Start by broadening $(a-b)^3$ in any manner girlfriend 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 various other side of the equality into the standard formula.

\beginalign* (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).\endalign*

This is one handy derivation. The sum of cubes have the right to be derived in a similar manner.

share
point out
monitor
edited Sep 4 "13 at 21:56
reply Sep 4 "13 in ~ 21:37 J. W. PerryJ. W. Perry
$\endgroup$
1
$\begingroup$
The 2nd equality is derived by first grouping terms in the middle expression and also then factoring the group terms:

\beginaligna^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\endalign

If the OP is wondering wherein the center expression came from in the first place, it"s type of a deus ex machina: all you"re doing is sticking 2 $0$"s between $a^3$ and $-b^3$ (the expressions $-a^2b+a^2b$ and also $-ab^2+ab^2$ room both obviously equal to $0$), yet when girlfriend do, lo and behold, the regrouping and factoring occupational their magic.

share
cite
monitor
reply Sep 4 "13 at 22:16 Barry CipraBarry Cipra
$\endgroup$
include a comment |
0
$\begingroup$
I am not certain I obtain your question. The critical formula is just obtained factoring a and also b. Because that example, the very first part

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

So as soon as you get the critical formula you have the right to take $(a-b)$ and obtain

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

and to acquire the second formula friend simply include and subtract the very same quantities, in specific $-a^2b+a^2b$ and also $ab^2-ab^2$.

re-publishing
mention
monitor
answered Sep 4 "13 in ~ 20:39 UmbertoUmberto
$\endgroup$
include a comment |

Thanks because that contributing solution to neurosoup.orgematics stack Exchange!

But avoid

Asking because that help, clarification, or responding to other answers.Making statements based on opinion; back them increase with recommendations or personal experience.

Use neurosoup.orgJax to format equations. Neurosoup.orgJax reference.

To find out more, check out our advice on writing an excellent answers.

See more: The I Have To Go My Planet Needs Me Gifs, Explore I Have To Go Now My Planet Needs Me Gifs

Draft saved

authorize up making use of Email and also Password
send

### Post together a guest

surname
email Required, yet never shown

### Post as a guest

name
email

Required, however never shown

## Not the price you're feather for? Browse various other questions tagged algebra-precalculus or asking your own question.

Featured on Meta
35 votes · comment · stats
associated
1
Relation in between "harmonic form" and also fourier series?
17
High institution algebra textbooks because that gifted students
2
Polynomial Factoring
8
far better way of identifying pairs of natural numbers the satisfy certain conditions?
1
Factoring equations?
1
just how to discover equation the parabola v diagonal directrix
2
Audio - Compute samples for a offered length with a start and also end tempo
1
hot Network inquiries an ext hot concerns

question feed

neurosoup.orgematics
company
stack Exchange Network
site style / logo © 2021 ridge Exchange Inc; user contributions license is granted under cc by-sa. Rev2021.9.9.40167

neurosoup.orgematics stack Exchange works finest with JavaScript allowed her privacy

By clicking “Accept all cookies”, you agree stack Exchange have the right to store cookie on your maker and disclose details in accordance through our Cookie Policy.