Factorization of $18(ab^2 + bc^2 + ca^2) - 12(a^2b + b^2c + c^2a) - 19abc$ I solved a following problem in my mathematics homework today:

Factorize $18(ab^2 + bc^2 + ca^2) - 12(a^2b + b^2c + c^2a) - 19abc$.

That's easy. I solved by the following way and it was same as sample answer of the book:

$\ \ \ \ 18(ab^2 + bc^2 + ca^2) - 12(a^2b + b^2c + c^2a) - 19abc$
  $=(-12b+18c)a^2 + (18b^2 - 19bc - 12c^2) + (18bc^2 - 12b^2c)$
  $=-6a^2 (2b-3c) + a(2b-3c)(9b+4c) - 6bc(2b-3c)$
  $=(2b-3c)(-6a^2 + 9ab + 4ac - 6bc)$
  $=(2b-3c)(2a-3b)(2c-3a)$.

(if there is a typo, please inform this)
Though the process took much times (highly implementing one), and I think it's too straight forward way.
The question and answer looks simple, so I think there must be more elegant and way, which has less calculation time.

So, I want to ask the simplier way to solve this.

Without loss of generality, assume $a, b=ax, c=axy$. Then:
$$18(ab^2 + bc^2 + ca^2) - 12(a^2b + b^2c + c^2a) - 19abc=$$
$$18a^3x(x + x^2y^2 + y) - 12a^3x(1 + x^2y + xy^2) - 19a^3x^2y=$$
$$a^3x[18x+18x^2y^2+18y-12-12x^2y-12xy^2-19xy]=$$
$$a^3x[6x(3-2xy)+6y(3-2xy)-(3-2xy)(4+9xy)]=$$
$$a^3x(3-2xy)(6x+6y-4-9xy)=$$
$$a^3x(3-2xy)(2-3y)(3x-2)=$$
$$(3a-2axy)(2ax-3axy)(3ax-2a)=$$
$$(3a-2c)(2b-3c)(3b-2a).$$
