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 $n=3$.
 A: 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*}
A: For cases of $n=5$ and $n=7$, let's have a look at the following identities
$$(x+y+z)^3 - (x^3+y^3+z^3) = 3(x+y)(x+z)(y+z)$$
$$(x+y+z)^5 - (x^5+y^5+z^5) = 5(x+y)(x+z)(y+z)(x^2+y^2+z^2+xy+xz+yz)$$
$$(x+y+z)^7 - (x^7+y^7+z^7) = 7(x+y)(x+z)(y+z)((x^2+y^2+z^2+xy+xz+yz)^2+xyz(x+y+z))$$
The three identities listed above are called Lame-type identities (see here).
Algebraically manipulating the second and third identities give;
Let
$$A = x+y+z,$$
$$B = x^2+y^2+z^2,$$
$$C = x^3+y^3+z^3,$$
$$E = x^5+y^5+z^5,$$
$$G = x^7+y^7+z^7,$$
then
\begin{equation}
6E=A^5-5BA^3+5CA^2+5BC \tag{1}
\end{equation}
\begin{equation}
36G=A^7+7CA^4-21B^2A^3+28C^2A+21B^2C \tag{2}
\end{equation}
By the way, I think they have nice applications. For example, if the values of $A, $ $B, $ and $E$ are given, $x $, $y $ and $z$ can be found with ease since $C$ can be generated from (1) and that, solving $A, $ $B, $ and $C$ simultaneously won't be difficult. Same thing goes for $A, $ $B, $ and $G$.
