Answer: Option C
$$\eqalign{
& {\text{ }}2x + \frac{2}{x} = 3 \cr
& \Rightarrow {\text{ }}x + \frac{1}{x} = \frac{3}{2} \cr
& {\text{Taking cube on both sides}} \cr
& \Rightarrow {\left( {x + \frac{1}{x}} \right)^3} = {\left( {\frac{3}{2}} \right)^3}{\text{ }} \cr
& x + \frac{1}{x} = a \cr
& {x^3} + \frac{1}{{{x^3}}} = {a^3} - 3a \cr
& \Rightarrow {x^3} + \frac{1}{{{x^3}}} + 3\left( {x + \frac{1}{x}} \right) = \frac{{27}}{8} \cr
& \Rightarrow {x^3} + \frac{1}{{{x^3}}} + 3 \times \frac{3}{2} = \frac{{27}}{8} \cr
& \Rightarrow {x^3} + \frac{1}{{{x^3}}} = \frac{{27}}{8} - \frac{9}{2} \cr
& \Rightarrow {x^3} + \frac{1}{{{x^3}}} = \frac{{ - 9}}{8} \cr
& \therefore {x^3} + \frac{1}{{{x^3}}} + 2 \cr
& = \frac{{ - 9}}{8} + 2 \cr
& = \frac{{ - 9 + 16}}{8} \cr
& = \frac{7}{8} \cr} $$