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The question is in the title.

This question is from "Algebra" by Gelfand.

My initial thought is that if $a$, $b$ and $c$ are $1$ or $-1$, then the polynomial evaluates to $0.$ So, maybe two of the factors will be $(a + b + c - 3)$ and $(a + b + c + 3)$. An alternative option that combines these two might be $a^{2} + b^{2} + c^{2} - 3$.

  1. Is the thought process correct here, and would trial and error be a good way to decide between the linear and the quadratic options I described above?
  2. As you can tell, I am largely doing guess work here. Is there a more systematic way of deciding what terms to add and subtract in orders to factor the polynomial?

Note: The factoring need not be done all the way to linear factors. All that is needed is a product of polynomials.

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  • $\begingroup$ A cryptic hint: it's the determinant of $\pmatrix{a&b&c\\c&a&b\\b&c&a}$. $\endgroup$ Commented Apr 20, 2018 at 16:26
  • $\begingroup$ Thanks for the hint, but I am trying not to rely on any knowledge not yet presented in the material. This is supposed to be an introductory algebra book. $\endgroup$
    – ski
    Commented Apr 20, 2018 at 16:27
  • $\begingroup$ This is a specific case of Newton-Gerard Identities. They are fairly basic to understand. $\endgroup$ Commented Nov 6, 2023 at 4:37

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I suggests that you use $(a+b)^3=a^3+b^3+3ab(a+b)\Rightarrow a^3+b^3=(a+b)^3-3ab(a+b)$ instead, you will need to use it twice like this:

$a^3+b^3+c^3-3abc$

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

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

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

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

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

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

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Since the given polynomial is homogeneous and symmetrical w.r.t. a,b,c there can be only one linear factor a+b+c. You can substitute -(b+c) for a and prove that the result is zero and verify that a factor is a+b+c. Since the polynomial is of 3rd degree the remaining factor should be of the form A(a² + b² + c²) +B(ab+bc+ca). Now by equating the coefficients or by substituting values for a,b,c it can be obtained A =1 , B = -1 .

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  • $\begingroup$ " Since the polynomial is of 3rd degree the remaining factor should be of the form A(a² + b² + c²) +B(ab+bc+ca)" How do you know this? $\endgroup$ Commented Jun 19, 2023 at 23:15
  • $\begingroup$ @SRobertJames since the given polynomial is cyclic homogeneous, all factors need to be of same kind. We have only two such polynomials of 2nd degree for 3 variables, therefore you have to use a combination of both and determine the constants A and B. $\endgroup$ Commented Jun 20, 2023 at 15:18
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Factor $a^3+b^3+c^3-3abc$ to a product of polynomials.

Think when $a=b=c$.

$$a^3+b^3+c^3-3abc=a^3+a^3+a^3-3a^3=0.$$

This only fits when $a=b=c$, but not when $a=b$ or $b=c$ or $c=a$.

So, you can think about $(a-b)^2+(b-c)^2+(c-a)^2$, which is $0$ if and only if $a=b=c$.

Therefore, you can reason $a^2+b^2+c^2-ab-bc-ca$ as the factor of $a^3+b^3+c^3-3abc$.

Also, think when $a=-b-c$.

$$a^3+b^3+c^3-3abc=(-b-c)^3+b^3+c^3-3(-b-c)bc \\ =-b^3-3b^2c-3bc^2-c^3+b^3+c^3+3(b+c)bc=0.$$

ISW, you can reason $a+b+c$ as the factor of $a^3+b^3+c^3-3abc.$

Since the degree of $a^2+b^2+c^2-ab-bc-ca$ is $2$, while $a+b+c$ has $1$, You can reason $(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=a^3+b^3+c^3-3abc$. checking this, you can find that you got the right one.

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  • $\begingroup$ How do you know that $(a^2+b^2+c^2−ab−bc−ca)$ can't be factored further? $\endgroup$ Commented Jun 19, 2023 at 23:16
  • $\begingroup$ $a^2+b^2+c^2-ab-bc-ca=a^2-(b+c)a+b^2-bc+c^2$. Then, It's sure that the term can't be factored with variable $a$, and it's symmetric for $a, b, c$. Therefore, there can't be more factorization for this term. $\endgroup$
    – RDK
    Commented Jun 22, 2023 at 16:26
  • $\begingroup$ Can you explain what you mean by "factored with variable $a$" and how you know that can't be done? $\endgroup$ Commented Jun 22, 2023 at 17:48
  • $\begingroup$ When we say that the polynomial is factored, it means that the polynomial can be expressed as $\left(k_na^n+k_{n-1}a^{n-1}+\cdots+k_1a^1+k_0a^0\right)P(a)$. Also, I showed that $a^2+b^2+c^2-ab-bc-ca$ is symmetric with $a, b, c$ and can be expressed as $a^2-(b+c)a+(b^2-bc+c^2)$. Using Vieta's formula, we get that $\alpha+\beta=b+c, \alpha\beta=b^2-bc+c^2$. Then we can say that $(\alpha-\beta)^2=(b+c)^2-4(b^2-bc+c^2)=-3b^2+6bc-3c^2=-3(b-c)^2\leq0$, which means that $b=c$. However, this polynomial should be factored for all $b, c \in \Bbb{R}$, so it can't be factored when $b \neq c$. $\endgroup$
    – RDK
    Commented Jun 23, 2023 at 10:39
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Hint: divide $$a^3+b^3+c^3-3abc$$ by $a+b+c$ the result is given by $$\left( c+a+b \right) \left( {a}^{2}-ab-ca+{b}^{2}-bc+{c}^{2} \right) $$

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    $\begingroup$ How is this a hint? It gives the entire solution. $\endgroup$
    – Arthur
    Commented Apr 20, 2018 at 16:27
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    $\begingroup$ What I am more interested in is how you arrive at that solution. What in the problem gives you the idea to factor $(c + a + b)$ ? $\endgroup$
    – ski
    Commented Apr 20, 2018 at 16:28
  • $\begingroup$ @ski It kindof looks like $(a+b+c)^3$, minus a few terms. Maybe one of those three factors of $a+b+c$ survived the removal of terms. More rigorously, your expression is homogenous (all terms have the same degree) and symmetric, which means any factor is probably homogenous and symmetric. $a+b+c$ is the only degree-1 symmetric, homogenous polynomial, so it's natural to try that first. $\endgroup$
    – Arthur
    Commented Apr 20, 2018 at 16:31
  • $\begingroup$ since by AM-GM gives $$\frac{a^3+b^3+c^3}{3}\geq abc$$ for $$a,b,c$$ nonnegative numbers $\endgroup$ Commented Apr 20, 2018 at 16:32
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The questioner is seeking motivation - seeking to know what would inspire a diligent beginning student (reading Gelfand) to come up with this factorization.

The best answer is that the student already knows the more basic factorization for the sum of two cubes:

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

The student is inspired to try to extend this in some way to the sum of three cubes.

So he tries to factor out the term $a+b+c$ from $a^3+b^3+c^3$. When he uses long division in the ordinary way, he determines that there is a remainder of $3abc$. Subtracting the remainder from the dividend, he obtains an exact factorization. This approach motivates (and in fact derives) the desired result.

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