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

Question

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.

Answer 1

You can make the followin.

Let $a=3$ and $b=2$.

Thus, our expression is equal to $0$.

Since our expression is homogenious,

we got a factor $2a-3b$ and since it's a cyclic expression, we get $$\sum_{cyc}(18a^2c-12a^2b)-19abc=A(2a-3b)(2b-3c)(2c-3a).$$

Now, let $a=b=c=1$.

We obtain $-1=-A$, which gives $A=1$ and we are done!

Answer 2

The expression is a homogeneous expression.

We can dehomogenize it by setting $b=1, c=1$. We get $ 6 a^2 - 13 a + 6 = (3 a - 2) (2 a - 3)$.

Rehomogenizing, this suggests that $(3 a - 2b) (2 a - 3c)$ or $(3 a - 2c) (2 a - 3b)$ is a factor.

Trying them out, we see that $(3 a - 2c) (2 a - 3b)$ is indeed a factor.

Answer 3

So that variables don't get lost amid constants, I recommend making the substitution $$b \to d \qquad c \to d \tag{1}$$ Then the polynomial reduces to $$6 a^2 d - 13 a d^2 + 6 d^3 = ( 2 a -3 d )( 3 a - 2 d ) d \tag{2}$$ From here, we can attempt to reinstate the polynomial's symmetry in the variables. It appears that taking $d \to b$ in the first factor, and $d\to c$ in the second factor is a pretty good guess, especially if we pull a negative out of the second factor ... $$(2 a - 3 d )( 3 a - 2 d) d \quad\to\quad -(2 a - 3 b)(2 c - 3 a) d \tag{3}$$ If there's any justice, the factorization would include $2b-3c$; such a factor is certainly plausible, since, under the substitution $(1)$, $2b-3c \to 2d - 3 d = -d$ (which incorporates that pesky negative). As it happens, the factorization really is $$(2a - 3 b)( 2b - 3 c)( 2 c - 3 a) \tag{4}$$ and we're done. $\square$

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Of course, there was some lucky guessing here. With $b$ and $c$ masked by $d$, it could have been that the $3d$ in $(2)$ expanded as $b+2c$, or $2b+c$, or $4b-c$, or, or, or ... . The $2d$ term is likewise ambiguous, as is the final $d$ term. Symmetry helps us discard these options: since the last factor of $(2)$ involves only $d$ ---that is, only $b$ and/or $c$--- we expect that each of the first two factors involves at most two variables, $a$ and either $b$ or $c$.

That said, even symmetry considerations fail to immediately rule-out the alternative guess of $d\to c$ in the first factor and $d\to b$ in the second ... $$(2 a - 3 d )( 3 a - 2 d) d \quad\to\quad -(2 a - 3 c)(2 b - 3 a) d \tag{$3^\prime$}$$ Here, the "justice" factor would be $2c - 3 b$, to provide a perfectly plausible unmasking of $-d$. So, we could just as easily have been looking at this candidate: $$(2 a - 3 c)(2 b - 3 a)(2 c - 3 b) \tag{$4^\prime$}$$ This one just happens to be wrong. (The $18$ and $-12$ coefficients attach to the wrong terms.)

Thus, it is not enough to make a guess that satisfies symmetry. Always check your answer.

Answer 4

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