1. ## a cubic

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

2. Hello, I got
$a^3+b^3+c^3-3abc = a(a^2-bc)+b(b^2-ac)+c(c^2-ab)=$

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

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

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

3. Originally Posted by pacman
Factor a^3 + b^3 + c^3 - 3abc , where to start with this?
You only have to remember it:
$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)$

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

5. Originally Posted by RobLikesBrunch
Simply memorizing mathematics without understanding is absurd and idiotic.
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 $x^3+y^3 = (x+y)(x^2+y^2 -xy)$.
So here is the other way how to factorize, and pacman you'll find another ways if you try.

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

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

$=(a+b+c)((a+b)^2+c^2-(a+b)c) -3ab(a+b+c) =$
$=(a+b+c)((a+b)^2+c^2-(a+b)c -3ab) =(a+b+c)(a^2+b^2+c^2-ab-bc-ac)$

6. 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.

7. 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 )

8. 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.