For which $n$ does $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}$ imply $\frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n}=\frac{1}{a^n+b^n+c^n}$ I'm having trouble finishing a problem on an old national competition.
As the title states, the question says asks:
Given $a,b,c \neq 0,a+b=c$ such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}$,
Find all integers $n$ such that $\frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n}=\frac{1}{a^n+b^n+c^n}$.
I know that if it is true for $n$, it is also true for $-n$ since $(a^n+b^n+c^n)(\frac{1}{a^n+b^n+c^n})=1$ and $(a^n+b^n+c^n)(\frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n})=1$ and the statment follows for $-n$ given that we can make the substitution for $n$. Thus I found out that $n=-1$ satisfies the equality.
I also know that it fails for all even $n$. If $n$ were even, $(a^n,b^n,c^n)$ would all be positive, and Titu's Lemma asserts that $\frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n}\geq \frac{9}{a^n+b^n+c^n}$. Thus $n$ must be odd.
I also observed that $(2,-1,1)$ was a solution, and from here noted that this particular solution satisfies all odd $n$.
I have been trying to prove that all other solutions satisfy all odd $n$ with no success.
Any help would be greatly appreciated. Thanks!

Lemma: if $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}$,then we have
$$\dfrac{1}{a^n}+\dfrac{1}{b^n}+\dfrac{1}{c^n}=\dfrac{1}{a^n+b^n+c^n}$$
where $n$ is odd.
proof:
$$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}
\Longrightarrow \dfrac{ab+bc+ac}{abc}=\dfrac{1}{a+b+c}$$
$$\Longrightarrow (a+b+c)(ab+bc+ac)=abc$$
since
$$(a+b+c)(ab+bc+ac)=(a+b)(b+c)(a+c)+abc\Longrightarrow (a+b)(b+c)(a+c)=0$$
WOLG, let $a=-b$,and Note $n$ is odd.so
$$a^n+b^n=0,\dfrac{1}{a^n}+\dfrac{1}{b^n}=0$$
so
$$\dfrac{1}{a^n}+\dfrac{1}{b^n}+\dfrac{1}{c^n}=\dfrac{1}{a^n+b^n+c^n}$$
