# Unclear math contest solution

I'm having difficulty understanding this solution to a problem in an old math contest. I understand everything up to the word "Hence". Can anybody explain this to me?

Question

• If $a^5=-3+9a-a^2$ and $b$ and $c$ satisfy the same equality, then we add and we get ... – mfl Oct 20 '17 at 2:35
• With which part of the solution were you having trouble? – Thomas Andrews Oct 20 '17 at 2:35
• Ohhh, mfl just made me understand it, thanks a lot! – S. Dob Oct 20 '17 at 2:40

The three roots satisfy the polynomial. So, from $$x^5 = \cdots = -3 + 9x - x^2$$ we get \begin{align*} a^5 &= -3 + 9a - a^2 \\ b^5 &= -3 + 9b - b^2 \\ c^5 &= -3 + 9c - c^2 \text{.} \end{align*} Summing these, we get $$a^5 + b^5 + c^5 = 3(-3) + 9(a+b+c) - (a^2 + b^2 + c^2) = \dots \text{.}$$
Alternatively: From Vieta's theorem: $$a+b+c=0, ab+bc+ac=-3, abc=-1.$$ We can find: $$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ac)=0^2-2(-3)=6.$$
From the equation: $$\begin{cases} a^3=3a-1 \\ b^3=3b-1 \\ c^3=3c-1 \end{cases} \Rightarrow a^3+b^3+c^3=3(a+b+c)-3=-3.$$ Multiply the two: $$(a^2+b^2+c^2)(a^3+b^3+c^3)=-18 \Rightarrow$$ $$a^5+b^5+c^5+(abc)^2(ab+bc+ac)=-18 \Rightarrow$$ $$a^5+b^5+c^5=-18-(-1)^2(-3)=-15.$$