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.

  • $\begingroup$ A cryptic hint: it's the determinant of $\pmatrix{a&b&c\\c&a&b\\b&c&a}$. $\endgroup$ 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
    Apr 20, 2018 at 16:27

5 Answers 5


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:









Hint: divide $$a^3+b^3+c^3-3abc$$ by $a+b+c$ the result is given by $$\left( c+a+b \right) \left( {a}^{2}-ab-ca+{b}^{2}-bc+{c}^{2} \right) $$

  • 3
    $\begingroup$ How is this a hint? It gives the entire solution. $\endgroup$
    – Arthur
    Apr 20, 2018 at 16:27
  • 1
    $\begingroup$ 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
    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
    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 $\endgroup$ Apr 20, 2018 at 16:32

Factor $a^3+b^3+c^3-3abc$ to a product of polynomials.

Think when $a=b=c$.


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.


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:


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.


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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