Given that: $2x+\frac{2}{x}=3$
$\implies 2(x+1/x)=3$
$\implies (x+\frac{1}{x})=\frac{3}{2}…….(i)$
cubing eq (i) we get:
$\implies x^3+\frac{1}{x^3}+3*x*\frac{1}{x}(x+\frac{1}{x})=\frac{27}{8}$
$\implies x^3+\frac{1}{x^3}+3*\frac{3}{2}=\frac{27}{8}$
$\implies x^3+\frac{1}{x^3}=\frac{27}{8}-\frac{9}{2}$
Adding $+2$ in both side we get:
$\implies x^3+\frac{1}{x^3}+2 =\frac{27}{8}-\frac{9}{2}+2$
$\implies x^3+\frac{1}{x^3}+2=\frac{27-36+16}{8}=\frac{7}{8}$
Option (B) is correct.