Understanding $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ by means of determinants. The identity comes from expanding the determinant
$$\begin{vmatrix}
a & b & c \\
c & a & b \\
b & c & a \\
\end{vmatrix}$$
in two ways.
The LHS comes from expanding the determinant by Sarrus' rule. The RHS comes from adding up all columns to the first, factoring (a+b+c), and then expanding the remaining determinant.

The derivation of the RHS is what I don't understand. I'm guessing that there is some elementary determinant operation at work that I'm unfamiliar with.

 A: In more detail, the thing that's being suggested to get the RHS is the following reasoning:
\begin{align}
 \begin{vmatrix}a & b & c \\ c & a & b \\ b & c & a\end{vmatrix} &= \begin{vmatrix}a+c & a+b & b+c \\ c & a & b \\ b & c & a\end{vmatrix} & \text{(Add row 2 to row 1)} \\
&= \begin{vmatrix}a+b+c & a+b+c & a+b+c \\ c& a & b \\ b & c & a\end{vmatrix} & \text{(Add row 3 to row 1)} \\
&= (a+b+c) \begin{vmatrix}1 & 1 & 1\\c & a & b\\ b & c & a\end{vmatrix} & \text{(Factor out $a+b+c$)} \\
&= (a+b+c) \left(\begin{vmatrix}a & b \\ c & a\end{vmatrix} - \begin{vmatrix}c & b \\ b & a\end{vmatrix} + \begin{vmatrix}c & a \\ b & c\end{vmatrix}\right) & \text{(Expand on $1^{\text{st}}$ row)} \\
&= (a+b+c)(a^2-bc - (ca-b^2) + c^2-ab) \\
&= (a+b+c)(a^2+b^2+c^2 - ab - bc - ca).
\end{align}
A: The matrix is a circulant matrix. Its eigenvectors are $(1,\zeta,\zeta^2)^t$ where $\zeta$ runs through the cube roots of unity. The
corresponding eigenvalues are $a+b\zeta+c\zeta^2$. The determinant is
thus
$$(a+b+c)(a+b\omega+c\omega^2)(a+b\omega^2+c\omega)$$
where $\omega=\exp(2\pi i/3)$. Pair off the last two factors to get your form.
As a follow-up problem, consider the determinant of
$$\pmatrix{a&b&c&d\\d&a&b&c\\c&d&a&b\\b&c&d&a}.$$
A: Add $\omega$ times the second row and $\omega^2$ times the third row to the first row to get
$\begin{align}
 \begin{vmatrix}a & b & c \\ c & a & b \\ b & c & a\end{vmatrix} &= \begin{vmatrix}a+c\omega+b\omega^2 & b+a\omega+c\omega^2  & c+b\omega + a\omega^2 \\ c & a & b \\ b & c & a\end{vmatrix} \\
&=\begin{vmatrix}a+c\omega+b\omega^2 & \omega(a+b\omega^2+c\omega)  & \omega^2(c\omega+b\omega^2 + a) \\ c & a & b \\ b & c & a\end{vmatrix} 
\end{align}$
and hence $a+c\omega+b\omega^2$ is a factor. Similar arguments can be used to show the other two factors in the proof given by Lord Shark the Unkniwn.
