If $$2x + \frac{2}{x} = 3{\text{,}}$$   then the value of $${x^3} + \frac{1}{{{x^3}}} + 2$$   is?

Solution(By Examveda Team)

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