If $${64^{x + 1}} = \frac{{64}}{{{4^x}}}{\text{,}}$$   then the value of x is?

$$\eqalign{ & {64^{x + 1}} = \frac{{64}}{{{4^x}}} \cr & \Rightarrow {\left( {{4^3}} \right)^{x + 1}} - \frac{{{4^3}}}{{{4^x}}} \cr & \Rightarrow {4^{3x + 3}} = {4^{3 - x}} \cr & \Rightarrow 3x + 3 = 3 - x \cr & \Rightarrow 4x = 0 \cr & \Rightarrow x = 0 \cr} $$