Answer: C
\(\alpha+\beta=-\cfrac{8}{3}\) and \(\alpha \beta=\cfrac{2}{3}\)
\(\therefore \cfrac{1}{\alpha}+\cfrac{1}{\beta}=\cfrac{\beta+\alpha}{\alpha\beta}= \cfrac{-\cfrac{8}{3}}{\cfrac{2}{3}}\)
\(\therefore =-\cfrac{8}{3}\times \cfrac{3}{2}=-4\)
\(\alpha+\beta=-\cfrac{8}{3}\) and \(\alpha \beta=\cfrac{2}{3}\)
\(\therefore \cfrac{1}{\alpha}+\cfrac{1}{\beta}=\cfrac{\beta+\alpha}{\alpha\beta}= \cfrac{-\cfrac{8}{3}}{\cfrac{2}{3}}\)
\(\therefore =-\cfrac{8}{3}\times \cfrac{3}{2}=-4\)