<s>[INST] How to factor $a^{3} + b^{3} + c^{3} - 3abc$ into a product of polynomials 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 $a^{2} + b^{2} + c^{2} - 3$. 


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*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?

*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.
 [/INST] 
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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