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 $n=3$.

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 $G$.
