Who came up with the identity $a^3+b^3+c^3-3abc=(a+b+c)\left[a^2+b^2+c^2-ab-bc-ca\right]$ 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 $a^3+b^3+c^3-3abc$ ?
 A: 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,
$$ (a^3 + b^3 + c^3 - 3abc)(p^3 + q^3 + r^3 - 3pqr) = x^3 + y^3 + z^3 - 3xyz $$
A: 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.
A: 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, $a,b,c\in\mathbb R$) $$\begin{align}S_n=&S_{n-1}(a+b+c)\\&-S_{n-2}(ab+bc+ca)\\&+S_{n-3}(abc)\end{align}$$
A: 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.
