Who came up with the identity $a^3+b^3+c^3-3abc=(a+b+c)\left[a^2+b^2+c^2-ab-bc-ca\right]$

Question

Though we can prove this it is not something that comes up intutively.

Our ancestors must have been interested in factorising $a^3+b^3+c^3$ but why find it for $a^3+b^3+c^3-3abc$ ?

Answer 1

Well, it is $$ \det( aI + b W + c W^2 ) $$ where $$ W = \left( \begin{array}{rrr} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \end{array} \right) $$

$$ \left( \begin{array}{rrr} a & b & c \\ c & a & b \\ b & c & a \end{array} \right) $$

Right. It follows that the polynomial is multiplicative: as $W^3 = I$ and $W^4 = W,$ we get $$ (aI + b W + c W^2)(pI + q W + r W^2) = xI + y W + z W^2, $$ where $$ x = ap + br + cq, $$ $$ y = aq + bp + cr, $$ $$ z = ar + bq + cp. $$ Then, by multiplication of determinants, $$ (a^3 + b^3 + c^3 - 3abc)(p^3 + q^3 + r^3 - 3pqr) = x^3 + y^3 + z^3 - 3xyz $$

Answer 2

This might better be a comment, but I'm posting it as an answer.

Let $S_n=a^n+b^n+c^n$.

More generally, we have ($n\ge 3$ is an integer, $a,b,c\in\mathbb R$) $$\begin{align}S_n=&S_{n-1}(a+b+c)\\&-S_{n-2}(ab+bc+ca)\\&+S_{n-3}(abc)\end{align}$$

Answer 3

People were interested already a long time ago in the Diophantine equation, $$ a^3+b^3+c^3-nabc=0, $$ for $n\ge 1$. For a modern treatment see here. The case $n=3$ inspired - perhaps- to look out for factorisations.

Answer 4

I don't know the history of this identity, but here is a derivation based on the properties of determinants.

$$\begin{vmatrix} a &c &b\\ b &a &c\\ c &b &a \end{vmatrix} = \begin{vmatrix} a+b+c &a+b+c &a+b+c\\ b &a &c\\ c &b &a \end{vmatrix} = (a+b+c) \begin{vmatrix} 1 &1 & 1\\ b &a &c\\ c &b &a \end{vmatrix} $$

Expanding the determinant on the left, we have $$a^3+b^3+c^3-3abc$$ and expanding the determinant on the right we have $$(a+b+c)(a^2+b^2+c^2-ab-ac-bc)$$ so these two expressions are equal.