<s>[INST] $a + b + c = 0$ Prove that $\frac{a^2 + b^2 + c^2}{2} * \frac{a^3 + b^3 + c^3}{3} = \frac{a^5 + b^5 + c^5}{5}$ The question says: $a + b + c = 0$
Prove that $$\frac{a^2 + b^2 + c^2}{2} * \frac{a^3 + b^3 + c^3}{3} = 
\frac{a^5 + b^5 + c^5}{5}$$
So I started with simplifying each part of expression

*

*$$ \frac{a^2 + b^2 + c^2}{2} = \frac{(a+b+c)^2 -2ab-2ac-2bc}{2} =
-(ab + ac + bc)$$
Because $a+b+c=0$


*$$ \frac{a^3+b^3+c^3}{3} = \frac{(a+b+c)^3 - 3a^2b-3ab^2-3a^2c-3ac^2-3b^2c-3bc^2-6abc}{3}
 = -(a^2b+ab^2+a^2c+ac^2+b^2c+bc^2 -2abc) $$
So, my question is, what can I do with the second part? Like $\frac{(a+b+c)^2(a+b+c)^3 - ...}{5}$
 [/INST] From $a+b+c=0$, we can replace every occurrence of $a$ with $-b-c$, turning this into a two variable problem.  Then, after simplifying the various expressions, we find
\begin{align*}
\frac{(-b-c)^2+b^2+c^2}{2} &= b^2+bc+c^2  \\
\frac{(-b-c)^3+b^3+c^3}{3} &= -b^2c-bc^2  \\
\frac{(-b-c)^5+b^5+c^5}{5} &= -b^4c -2b^3c^2 -2b^2c^3 -bc^4
\end{align*}
It is straightforward to show the product of the first two is the third.
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