How to Factor a^3 + b^3 + c^3 – 3abc into Polynomials

algebra-precalculus

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.

Best Answer

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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