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

This has been proved How to show that $A^3+B^3+C^3 - 3ABC = (A+B+C)(A+B\omega+C\omega^2)(A+B\omega^2+C\omega)$ indirectly? and Find $a^3+b^3+c^3-3abc$ (binomial theorem) using algebra.

But is it possible to prove it using the factorization of homogeneous symmetric polynomials ?

Attempt

$a^3+b^3+c^3-3abc$ is a homogeneous symmetric polynomial of degree 3 so the factors are to be either 3 one degree polynomials or two one-degree and one two degree polynomial, all symmetric. Elementary symmetric polynomials of 3 variables are $1, a+b+c, ab+bc+ca,abc$. Clearly $abc$ is not a factor.

SOORAJ SOMAN
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2 Answers2

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Let $f(a,b,c)=a^3+b^3+c^3-3abc$.

\begin{align*} f(-b-c,b,c)&=-(b+c)^3+b^3+c^3+3(b+c)bc\\ &=(b+c)(-b^2-2bc-c^2+b^2-bc+c^2+3bc)\\ &=0 \end{align*}

By Factor Theorem, $f(a,b,c)$ is divisible by $a+b+c$.

Therefore, $f(a,b,c)=(a+b+c)[M(a^2+b^2+c^2)+N(ab+bc+ca)]$ for some $M,N\in\mathbb{R}$.

$M=f(1,0,0)=1$

$2(2M+N)=f(1,1,0)=2$

So, $N=-1$.

CY Aries
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  • I don't see what is implied by calculating $f(-b-c,b,c)=0$, could someone help me to understand this, thanks – Dabed Jul 02 '19 at 23:07
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    It is the Factor Theorem. If $g(x)$ is a polynomial and $g(k)=0$, then $g(x)$ is divisible by $x-k$. Now if we fix $b$, $c$ and consider the polynomial $g(x)=f(x,b,c)$, then $f(-b-c,b,c)=0$ is just $g(-b-c)=0$ and therefore $g(x)$ is divisible by $x+b+c$. – CY Aries Jul 03 '19 at 00:21
  • I was doing just $f(a+b+c,a+b+c,a+b+c)=0$ but that is not how the Factor Theorem goes, $g(x)=f(x,b,c)$ clarifies everything thank you very much – Dabed Jul 03 '19 at 00:37
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$$ \begin{split} a^3+b^3+c^3-3abc &=a^3+3a^2b+3ab^2+b^3+c^3-3a^2b-3ab^2-3abc\\ &=(a+b)^3+c^3-3ab(a+b+c)\\ &=(a+b+c)((a+b)^2-(a+b)c+c^2-3ab)\\ &=(a+b+c)(a^2+b^2+c^2-ab-ac-bc). \end{split} $$

Daniele Tampieri
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Michael Rozenberg
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