If $$\left(x \sqrt{x}\right)^{x^{2}} = (x)^{x\sqrt{x}}$$, and if $$x ∉\left\{0,1\right\}$$, then $$x =$$

A

$$\frac{2}{3}$$

B

$$\frac{3}{2}$$

C

$$\frac{4}{9}$$

D

$$\frac{9}{4}$$