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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 $\pmatrix{a&b&c\\c&a&b\\b&c&a}$. $\endgroup$ – Lord Shark the Unknown Apr 20 '18 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 '18 at 16:27
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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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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 Apr 20 '18 at 16:27
  • $\begingroup$ 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 '18 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 '18 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 $\endgroup$ – Dr. Sonnhard Graubner Apr 20 '18 at 16:32

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