If x is real, $$x + \frac{1}{x} \ne 0$$   and $${x^3}{\text{ + }}\frac{1}{{{x^3}}} = 0{\text{,}}$$   then the value of $${\left( {x + \frac{1}{x}} \right)^4}\,{\text{is?}}$$

Solution(By Examveda Team)

$$\eqalign{ & {x^3}{\text{ + }}\frac{1}{{{x^3}}} = 0{\text{ }} \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^3} - 3 \times x \times \frac{1}{x}\left( {x + \frac{1}{x}} \right) = 0 \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^3} - 3\left( {x + \frac{1}{x}} \right) = 0 \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^3} = 3\left( {x + \frac{1}{x}} \right) \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^2} = 3 \cr & \,\,\,\,\left( {{\text{Squaring both sides}}} \right) \cr & \Rightarrow {\left[ {{{\left( {x + \frac{1}{x}} \right)}^2}} \right]^2} = {\left( 3 \right)^2} \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^4} = 9 \cr} $$