If x is real, x + 1/x 0 and x^3 + 1x^3 = 0 , then the value of ( x + 1/x )^4\, is?

Answer: Option B

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