[A] $-\dfrac{K}{L}$.
[B] $\dfrac{H}{L}$.
[C] $-\dfrac{H}{L}$.
[D] $\dfrac{K}{L}$.
Solução
De acordo com as relações de Girard, onde no polinômio $p(x)=x^{3}+Hx^{2}+Kx+L$, $a=1, b=H, c=K$ e $d=L$: $$p_{1}+p_{2}+p_{3}=\dfrac{-b}{a}=\dfrac{-H}{1}=-H$$ $$p_{1} \cdot p_{2}+p_{1} \cdot p_{3}+p_{2} \cdot p_{3}=\dfrac{c}{a}=\dfrac{K}{1}=K$$ $$p_{1} \cdot p_{2} \cdot p_{3}=\dfrac{-d}{a}=\dfrac{-L}{1}=-L$$ Logo, a soma dos inversos multiplicativos, $S=\dfrac{1}{p_{1}}+\dfrac{1}{p_{2}}+\dfrac{1}{p_{3}}$ é igual a $$S=\dfrac{1}{p_{1}}+\dfrac{1}{p_{2}}+\dfrac{1}{p_{3}}$$ $$S=\dfrac{p_{1} \cdot p_{2}+p_{1} \cdot p_{3}+p_{2} \cdot p_{3}}{p_{1} \cdot p_{2} \cdot p_{3}}$$ $$S=\dfrac{K}{-L}=-\dfrac{K}{L}$$
Resposta: Item A