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

Answer


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:
$$(a^2+b^2+c^2)(a^3+b^3+c^3)=-18 \Rightarrow$$
$$a^5+b^5+c^5+(abc)^2(ab+bc+ac)=-18 \Rightarrow$$
$$a^5+b^5+c^5=-18-(-1)^2(-3)=-15.$$
