2003 AMC 10A Problem 9

Work from the inside. The innermost piece is
$$\sqrt[3]{x \sqrt{x}} = \sqrt[3]{x^{1} \cdot x^{1/2}} = \sqrt[3]{x^{3/2}} = x^{1/2} = \sqrt{x}.$$
So the innermost cube root simply returns $\sqrt{x}$. The next layer is then again $\sqrt[3]{x\sqrt{x}}$, which is again $\sqrt{x}$, and the outermost layer is the same expression one more time.

Each layer reproduces $\sqrt{x}$, so the entire expression equals $\sqrt{x}$.

The answer is $\boxed{\textbf{(A)}\ \sqrt{x}}$.

Nested radicals are easiest as fractional exponents, evaluated inside-out. Here $\sqrt{x}$ is a fixed point of the map $t \mapsto \sqrt[3]{x t}$: once one layer produces $\sqrt{x}$, every further layer does too. Spotting a fixed point turns a three-layer computation into one line.

Plug in $x = 64$: $\sqrt{64} = 8$, then $\sqrt[3]{64 \cdot 8} = \sqrt[3]{512} = 8$, and the same happens twice more. The result is $8 = \sqrt{64}$, matching only choice (A).

• Trying to track the exponent through all three layers at once and mishandling the fractions, ending at an exponent like 1/54 or 80/81.
• Multiplying exponents where they should be added ($x \cdot x^{1/2}$ is $x^{3/2}$, not $x^{1/2}$).