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! 


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.
