How to factor $a^{3} + b^{3} + c^{3} - 3abc$ into a product of polynomials

Question

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.

Answer 1

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:

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

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

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

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

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

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

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

Answer 2

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

Answer 3

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.

Answer 4

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 .

Answer 5

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.