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?
 A: From the original equation, we get:
$$abc=(a+b+c)(ab+bc+ca)$$
which is equivalent to 
$$a^2(b+c)+bc(b+c)+ab(b+c)+ca(b+c)=0$$
$$\implies(b+c)(a+c)(a+b)=0$$
Then, obviously any one of the following must hold:
$$a=-b\\b=-c\\c=-a$$
In any case we can prove the equation $$\frac{1}{a^n+b^n+c^n}=\frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n}$$
with odd $n$. 
Since if we take $a=-b$
we get 
$$\frac{1}{(-b)^n+b^n+c^n}=\frac{1}{(-b)^n}+\frac{1}{b^n}+\frac{1}{c^n}$$
which is equivalent to
$$\frac{1}{c^n}=\frac{1}{c^n}$$
and this is true.....
A: Hint 


*

*At the first step, show that $(a+b)(a+c)(b+c)=0$ 

*Next, show that two of the three numbers are opposite.

A: 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 $$\left(a^{2k+1}+b^{2k+1}\right)\left(a^{2k+1}+c^{2k+1}\right)\left(b^{2k+1}+c^{2k+1}\right)=0.$$
A: This method of infinite ascent is probably wrong.. but...
Suppose none of $a = -b, b = -c, c = -a$ is true. 
Then multiplying both sides by $abc$ and inverting we get:
$\frac{a+b+c}{abc} = \frac{1}{ab + bc + ca} \implies \frac{1}{ab + bc + ca}=  \frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}$
We can now write $x = ab, y = bc, z = ca$ and do this all over again which is probably absurd (needs better logic here). 
