Find the value of $\frac{S_{5}S_{2}}{S_{7}}$ If $a$, $b$, $c$ $\in \mathbb R$, we define $S_{k}=\frac{a^k+b^k+c^k}{k}$ (where $k$ is a non-negative integer). Given that $S_{1}=0$, find the value of $$\frac{S_{5}S_{2}}{S_{7}}$$  

I tried:
  We are given that $S_{1}=0$ i.e. $a + b + c=0$ $\implies a^3+b^3+c^3=3abc$ or $S_{3}=abc$. Similarly, for getting a relation for $S_{2}$ I squared the given condition. However, for higher powers, finding a condition becomes tedious. Even after raising the given equation to power $5$ and $7$ and then substituting for $S_{5}$ and $S_{2}$, I'm not arriving at a particular answer. However, I think that the fact that the degree of numerator and denominator in $\frac{S_{5}S_{2}}{S_{7}}$ is equal can somehow be used but I'm not getting it. Please Help!
  Thanks! 

 A: Observe that
$$a^{n+3}+b^{n+3}+c^{n+3}=\\(a+b+c)(a^{n+2}+b^{n+2}+c^{n+2})-(ab+ac+bc)(a^{n+1}+b^{n+1}+c^{n+1})+abc(a^n+b^n+c^n)$$
So since $a+b+c=0$ we have
$$a^2+b^2+c^2=-2(ab+ac+bc)\\ a^3+b^3+c^3=3abc\\
 a^4+b^4+c^4=2(ab+ac+bc)^2\\
 a^5+b^5+c^5=-5abc(ab+ac+bc)\\
a^6+b^6+c^6=-2(ab+ac+bc)^3+3(abc)^2\\
a^7+b^7+c^7=7abc(ab+ac+bc)^2$$
From where
$S_5S_2-S_7=\frac{1}{10}10abc(ab+ac+bc)^2-\frac{1}{7}7abc(ab+ac+bc)^2=0$
A: My solution might not the easiest but it works. From the initial relation you get that $c=-(a+b)$ so:
$$S_k=\frac{a^k+b^k+(-1)^k(a+b)^k}{k}$$
So when k is odd you don't have the terms of order k. With the values of k provided you get:
$$\frac{7}{10} \left[ \frac{-10(a^4b + 2a^3b^2 + 2a^2b^3 + ab^4)(a^2 + b^2 + ab)}{-7(a^6b + 3a^5b^2+5a^4b^3 + 5 a^3b^4 + 3a^2b^5 +ab^6)} \right] = \frac{(a^4b + 2a^3b^2 + 2a^2b^3 + ab^4)(a^2 + b^2 + ab)}{(a^6b + 3a^5b^2+5a^4b^3 + 5 a^3b^4 + 3a^2b^5 +ab^6)}$$
Rearranging terms you find that the numerator and denominator are equal so:
$$\frac{S_5 S_2}{S_7} = 1$$
Probably there is an easier way to get it.
