2003 AMC 10A Problem 9

Working outward, $x\sqrt{x} = x^{\frac{3}{2}},$ and its cube root is $x^{\frac{1}{2}}.$ Then $x \cdot x^{\frac{1}{2}} = x^{\frac{3}{2}},$ whose cube root is again $x^{\frac{1}{2}}.$ Repeating once more, $x \cdot x^{\frac{1}{2}} = x^{\frac{3}{2}},$ whose cube root is $x^{\frac{1}{2}} = \sqrt{x}.$ Thus, the correct answer is A.