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

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- September 1st 2009, 07:28 AMpacmana cubic
Factor a^3 + b^3 + c^3 - 3abc , where to start with this?

- September 1st 2009, 08:13 AMTaluivren
Hello, I got

- September 1st 2009, 08:14 AMynj
- September 1st 2009, 09:38 AMRobLikesBrunch
- September 1st 2009, 11:27 AMTaluivren
(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 .

So here is the other way how to factorize, and pacman you'll find another ways if you try.

[now we make use of ]

- September 1st 2009, 06:11 PMpacman
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)

- September 2nd 2009, 12:17 AMTaluivren
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))

- September 2nd 2009, 02:03 AMpacman
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.