Can the formula for the cubes $a^3 + b^3 + c^3 - 3abc$ be generalized for powers other than 3? I recently learnt out about this formula:
$$a^3 + b^3 + c^3 - 3abc = (a + b + c) (a^2 + b^2 + c^2 - ab - ac - bc)$$
Is there a way of generalizing for powers other than $3$, i.e.
$$a^n + b^n + c^n + \mathop{???} = \mathop{???} $$
where the RHS is in terms of $(a^{n-1} + b^{n-1} + c^{n-1})$, $(a^{n-2} + b^{n-2} + c^{n-2})$, ... , $(a^{1} + b^{1} + c^{1})$, just like the formula for $n=3$.

Too long for a comment:
One possible generalization is the circulant determinant, which tells that
$$ \det \begin{pmatrix}
a_1 & a_2 & a_3 & \cdots & a_n \\
a_n & a_1 & a_2 & \cdots & a_{n-1} \\
a_{n-1} & a_n & a_1 & \cdots & a_{n-2} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
a_2 & a_3 & a_4 & \cdots & a_1
\end{pmatrix}
= \prod_{\omega \mathop{:} \omega^n = 1} (a_1 + a_2 \omega + a_3 \omega^2 + \cdot + a_n \omega^{n-1}). $$
When $n = 2$, this gives
$$ a^2 - b^2 = \det\begin{pmatrix} a & b \\ b & a \end{pmatrix} = (a + b)(a - b). $$
When $n = 3$, then with any $\omega$ satisfying $\omega^2 + \omega + 1 = 0$,
\begin{align*}
a^3 + b^3 + c^3 - 3abc
&= \det\begin{pmatrix} a & b & c \\ c & a & b \\ b & c & a \end{pmatrix} \\
&= (a + b + c)(a + b\omega + c\omega^2)(a + b\omega^2 + c\omega) \\
&= (a + b + c)(a^2 + b^2 + c^2 + (\omega + \omega^2)(ab + bc + ca)) \\
&= (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca).
\end{align*}
Unfortunately, the formula quickly goes out of control even starting at $n = 4$:
\begin{gather*}
a^4 - b^4 + c^4 - d^4 - 2a^2c^2 + 2b^2d^2 - 4da^2b + 4ab^2c - 4bc^2d + 4cd^2a\\
= (a+b+c+d)(a-b+c-d)(a^2 + b^2 + c^2 + d^2 - 2ac - 2bd)
\end{gather*}
