<s>[INST] 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 $a^3+b^3+c^3-3abc$ ?
 [/INST] 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,
$$ (a^3 + b^3 + c^3 - 3abc)(p^3 + q^3 + r^3 - 3pqr) = x^3 + y^3 + z^3 - 3xyz $$
</s>