0

How can I factor the expression below only using basic factoring techniques without symmetric polynomials? thank you in advance.

$$a^2b+a^2c+ab^2+b^2c+ac^2+bc^2−a^3−b^3−c^3−2abc$$

I had found the question in another topic, but the answer doesn't quite explain how to make a more basic form [factrorize $-a^3+a^2b+a^2c+ab^2-2abc+ac^2-b^3+b^2c+bc^2-c^3$ ].

Helen
  • 455

1 Answers1

0

If you "know" that there is a nice factorization, then because it is symmetric in 3 variables and has degree 3, it can only be one of the following (or their multiples):

$$ \prod (a+b-c), (a+b+c)^3 , (a+b+c)\left( k(a^2+b^2+c^2) + l (ab+bc+ca)\right) $$

We can then expand all of these and compare.

As related examples, try to factorize the following knowing that a nice factorization exists:

  1. $a^3+b^3+c^3 - 3abc$
  2. $a^4+b^4+c^4 + a^2b^2 + b^2c^2 + c^2a^2 -2abc(a+b+c)$
Calvin Lin
  • 68,864
  • That´s cheating, because you first use the fact that nice factorization exists. –  Nov 07 '19 at 00:05
  • That's my point (if you can't tell from the first line). – Calvin Lin Nov 07 '19 at 00:06
  • I know, but it can also be done differently. –  Nov 07 '19 at 00:09
  • Of course, so add in your approach. I was thinking of doing the "remainder-factor" way, but that also seems to be an after-the-fact justification. – Calvin Lin Nov 07 '19 at 00:11
  • see https://math.stackexchange.com/questions/2690926/determination-of-the-algebraic-image-of-a-polynomial-morphism/2694033#2694033 about less nice factors (of a homogeneous ternary cubic) and example: https://math.stackexchange.com/questions/2678486/how-to-show-that-if-x-y-z-are-rational-numbers-satisfying-x-y-z3-9/2678675#2678675 – Will Jagy Nov 07 '19 at 00:31