2008/09 AHSMC Part 1 Problem 14

Let $a$, $b$ and $c$ be the roots. Then $a + b + c = 0$ and $bc + ca + ab = -3$. It follows that $a^2 + b^2 + c^2 = (a+b+c)^2 - 2(bc+ca+ab) = 6$. Since $x^3 = 3x - 1$, we have $$x^5 = x^2(3x-1) = 3(3x-1) - x^2 = -3 + 9x - x^2.$$ Hence $a^5 + b^5 + c^5 = -9 + 9(a+b+c) - (a^2+b^2+c^2) = -15$. The answer is (A).