# a cubic

• Sep 1st 2009, 07:28 AM
pacman
a cubic
Factor a^3 + b^3 + c^3 - 3abc , where to start with this?
• Sep 1st 2009, 08:13 AM
Taluivren
Hello, I got
\$\displaystyle a^3+b^3+c^3-3abc = a(a^2-bc)+b(b^2-ac)+c(c^2-ab)= \$

\$\displaystyle =a(a^2-bc -ac -ab +b^2 +c^2)-a(-ac -ab +b^2 +c^2)+\$
\$\displaystyle +b(b^2-ac-bc-ab+a^2+c^2)-b(-bc-ab+a^2+c^2)+\$
\$\displaystyle +c(c^2-ab-ac-bc+a^2+b^2)-c(-ac-bc+a^2+b^2)=\$

\$\displaystyle =(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\$
\$\displaystyle +a^2c+a^2b-ab^2-ac^2 +b^2c+ab^2-a^2b-bc^2 +ac^2+bc^2-a^2c-b^2c=\$

\$\displaystyle =(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\$
• Sep 1st 2009, 08:14 AM
ynj
Quote:

Originally Posted by pacman
Factor a^3 + b^3 + c^3 - 3abc , where to start with this?

You only have to remember it:
\$\displaystyle a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)\$
• Sep 1st 2009, 09:38 AM
RobLikesBrunch
Quote:

Originally Posted by ynj
You only have to remember it:
\$\displaystyle a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)\$

Simply memorizing mathematics without understanding is absurd and idiotic. It is much better to not memorize a certain formula and derive it each time...even if it takes you longer.
• Sep 1st 2009, 11:27 AM
Taluivren
Quote:

Originally Posted by RobLikesBrunch
Simply memorizing mathematics without understanding is absurd and idiotic.

(Giggle) agree with you, but some things are good to know. I don't remember this formula, but what initially helped me to factor the expression was the formula \$\displaystyle x^3+y^3 = (x+y)(x^2+y^2 -xy)\$.
So here is the other way how to factorize, and pacman you'll find another ways if you try.

\$\displaystyle a^3+b^3+c^3-3abc = a^3+b^3+c^3-3abc +3ab(a+b) - 3ab(a+b)=\$
\$\displaystyle = (a+b)^3 +c^3 - 3abc -3ab(a+b) = (a+b)^3 +c^3 -3ab(a+b+c) =\$

[now we make use of \$\displaystyle x^3+y^3 = (x+y)(x^2+y^2 -xy)\$]

\$\displaystyle =(a+b+c)((a+b)^2+c^2-(a+b)c) -3ab(a+b+c) =\$
\$\displaystyle =(a+b+c)((a+b)^2+c^2-(a+b)c -3ab) =(a+b+c)(a^2+b^2+c^2-ab-bc-ac)\$
• Sep 1st 2009, 06:11 PM
pacman
I appreciate all you guys, but the hard thing about this is that how did you figure out the sequence of steps that leads to the identity . . . . that one alone bogged me down. Thanks for the efforts to help. (Clapping)(Clapping)(Clapping)(Clapping)(Clapping)
• Sep 2nd 2009, 12:17 AM
Taluivren
Quote:

Originally Posted by pacman
I appreciate all you guys, but the hard thing about this is that how did you figure out the sequence of steps that leads to the identity . . . . that one alone bogged me down.

for me, it is about messing around with symbols using identities i already know and recognizing patterns and symmetries. Maybe we should start to learn from computers, i suppose they are better in this than humans. Can anybody explain how computers do it? (i mean how they factorize (Wink))
• Sep 2nd 2009, 02:03 AM
pacman
i found this from the web, now i wonder why it is difficult to factor

The identity would probably be known to Lagrange from his extensive study of algebraic equations If w is a primitive cubic root of unity then a^3 + b^3 + c^3 - 3abc is the constant term of the polynomial satisfied by a +bw +cww
a^3 + b^3 + c^3 - 3abc = (a+b+c)(a +bw +cww)(a +bww +cw).

This and other similar identities occur when symmetrical functions of the roots of polynomial
equations are calculated I know Newton studied symmetric functions of roots. He may have been aware of this identity.

The (a +bw +cww) is a special form of Eisenstein cubic integrers and a^3 + b^3 + c^3 - 3abc ist its norm. thus, a^3 + b^3 + c^3 - 3abc can also be referrred to a a termary cubic form.