$P(1) = 0$ $\implies$ $1 + a - 13 + 12 = 0$ $\implies$ $a = 0$ . Assim: $P(x) = x^3 - 13x + 12 = (x-1)(x^2 + bx + c)$, logo:$$x^3 - 13x + 12 = x^3 + x^2(b-1) + x(c-b) - c$$Conclui-se que $b = 1$ e $c = -12$ $\implies$ $P(x) = (x-1)(x^2 + x - 12)$$$P(x) = (x-1)(x+4)(x-3)$$Raízes: $\{-4,1,3\}$ $\implies$ $\boxed{\text{Soma das raízes:} -4+1+3 = 0}$$$\text{Alternativa } \mathbb{(D)}$$