If x - 3/x = 6,\,x 0, then the value of x^4 - 27x^2x^2 - 3x - 3

Answer: Option D

$$\begin{array}{rl}& \frac{x^4-\frac{27}{x^2}}{x^2-3x-3}\\ & =\frac{x\left(x^3-\frac{27}{x^3}\right)}{x\left(x-3-\frac{3}{x}\right)}.....\left(\text{i}\right)\\ & x^3-\frac{27}{x^3}\\ & =6^3+3\times 3\times 6\\ & =216+54\\ & =270\\ & \frac{x^3-\frac{27}{x^3}}{x-\frac{3}{x}-3}=\frac{270}{6-3}=90\end{array}$$