For which $n$ does $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}$ imply $\frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n}=\frac{1}{a^n+b^n+c^n}$
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 $$\dfrac{1}{a^n}+\dfrac{1}{b^n}+\dfrac{1}{c^n}=\dfrac{1}{a^n+b^n+c^n}$$