Show that $\frac{1}{a^n+b^n+c^n} = \frac{1}{a^n} + \frac{1}{b^n} + \frac{1}{c^n}$ Let $n \in N, n=2k+1, and \text{ } \frac{1}{a+b+c} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$.

Show that $$\frac{1}{a^n+b^n+c^n} = \frac{1}{a^n} + \frac{1}{b^n} + \frac{1}{c^n}$$

I have tried, but I don't get anything. Can you please give me a hint?

The condition gives $$(a+b+c)(ab+ac+bc)=abc$$ or
$$\sum_{cyc}(a^2b+a^2c+abc)=abc$$ or
$$\sum_{cyc}\left(a^2b+a^2c+\frac{2}{3}abc\right)=0$$ or
$$\prod_{cyc}(a+b)=0.$$
We need to prove that
$$\prod_{cyc}(a^n+b^n)=0$$ or $$\prod_{cyc}(a+b)\prod_{cyc}\left(a^{2k}-a^{2k-1}b+...+b^{2k}\right)=0,$$ which is obvious.
We can use also the following reasoning.
Since the condition is true for $a=-b$ and it's symmetric and degree $3$, we obtain that the condition is equal to $$(a+b)(a+c)(b+c)=0$$ and we need to prove that $$\left(a^{2k+1}+b^{2k+1}\right)\left(a^{2k+1}+c^{2k+1}\right)\left(b^{2k+1}+c^{2k+1}\right)=0.$$
