I recently acquired interested in neurosoup.orgematics after having slacked via it in highinstitution. Therefore I picked up the book "Algebra" by I.M. Gelfand and also A.Shen

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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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\$.

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answered Sep 4 "13 at 20:39 UmbertoUmberto
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