If x is numerically so small so that $x^2$ and higher powers of x can be neglected, then $\bigg(1+ \large\frac{2x}{3}\bigg)^{\large\frac{3}{2}}$$.(32+5x)^{\large\frac{-1}{5}}$ is approximately equal to :

Question

$$\begin {array} {1 1} (1)\;\frac{32+31x}{64} & \quad (2)\;\frac{31+32x}{64} \\ (3)\;\frac{31-32x}{64} & \quad (4)\;\frac{1-2x}{64} \end {array}$$