$a + b + c = 0$ Prove that $\frac{a^2 + b^2 + c^2}{2} * \frac{a^3 + b^3 + c^3}{3} = \frac{a^5 + b^5 + c^5}{5}$

Question

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

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

2. $$ \frac{a^3+b^3+c^3}{3} = \frac{(a+b+c)^3 - 3a^2b-3ab^2-3a^2c-3ac^2-3b^2c-3bc^2-6abc}{3} = -(a^2b+ab^2+a^2c+ac^2+b^2c+bc^2 -2abc) $$

So, my question is, what can I do with the second part? Like $\frac{(a+b+c)^2(a+b+c)^3 - ...}{5}$

Answer 1

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.