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?

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edited Sep 7 "13 at 10:58
asked Sep 4 "13 at 20:30

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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 $n$.

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answer Sep 4 "13 in ~ 20:36

Tyler HoldenTyler Holden
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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.

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edited Sep 4 "13 at 21:56
reply Sep 4 "13 in ~ 21:37

J. W. PerryJ. W. Perry
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The 2nd equality is derived by first grouping terms in the middle expression and also then factoring the group terms:


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.

reply Sep 4 "13 at 22:16

Barry CipraBarry Cipra
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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


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

Does this answer her question?

answered Sep 4 "13 in ~ 20:39

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