How to prove this equation? $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = \frac{1}{a+b+c}$ Suppose $a, b$, and $c$ are nonzero real numbers which satisfy the equation: $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = \frac{1}{a+b+c}$
Prove: if $n$ is an odd integer, then $a^n + b^n + c^n=(a+b+c)^n$
I was thinking of relating this to natural logs and derivatives somehow but then i'm not even sure how that would help.
 A: Consider the expression
$$\frac{1}{x}+\frac{1}{b}+\frac{1}{c} = \frac{1}{x+b+c}.$$
This can be simplified to 
$$(x+b+c)(x(b+c)+bc) - x(bc)=0.$$
This is a second degree polynomial in $x$. If we set $x=-b$, then this satisfies the given equation, hence $x+b$ is a factor of this polynomial. By symmetry $x+c$ is also a factor of this polynomial. Thus for some constant $K$ we get
$$(x+b+c)(x(b+c)+bc) - x(bc)= K(x+b)(x+c).$$
By direct comparison the constant $K$ turns out to be $b+c$. Thus 
$$(x+b+c)(x(b+c)+bc) - x(bc)= (b+c)(x+b)(x+c).$$
Said differently we have
$$(a+b+c)(a(b+c)+bc) - a(bc)= (b+c)(a+b)(a+c)=0.$$
Thus 
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = \frac{1}{a+b+c}.$$ 
is equivalent to 
$$(a+b)(b+c)(c+a)=0.$$
Thus at least one of the following must be true 
$$a=-b \quad \text{ or } \quad b=-c \quad \text{ or } \quad c=-a.$$
Now we can use the fact that $n$ is odd to claim that 
$$a^n + b^n + c^n=(a+b+c)^n.$$
A: Note that:
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = \frac{1}{a+b+c}$$
$$\frac{ab+ac+bc}{abc}=\frac{1}{a+b+c}$$
$$(ab+ac+bc)(a+b+c)-abc=0$$
$$(a+b)(b+c)(c+a)=0$$
then $a=-b$, $a=-c$ or $b=-c$. 
If $a=-b$, how $n$ is odd, then 
$$a^n + b^n + c^n=(-b)^n + b^n + c^n=c^n$$ and $$(a+b+c)^n=(-b+b+c)^n=c^n$$
you continue...
