2003 AMC 10A Problem 9

Nested roots are easiest from the inside out, so start with the innermost piece, $\sqrt[3]{x\sqrt{x}}$.

1. Write $x$ as $\sqrt{x} \cdot \sqrt{x}$. Then
$$x\sqrt{x} = \sqrt{x} \cdot \sqrt{x} \cdot \sqrt{x} = \left(\sqrt{x}\right)^3.$$
2. The cube root undoes the cube, so
$$\sqrt[3]{x\sqrt{x}} = \sqrt{x}.$$
3. Replace the innermost piece by $\sqrt{x}$. The whole expression becomes
$$\sqrt[3]{x\sqrt[3]{x\sqrt{x}}}.$$
Its inside is again $\sqrt[3]{x\sqrt{x}} = \sqrt{x}$, so the expression shrinks to $\sqrt[3]{x\sqrt{x}}$.
4. That is $\sqrt{x}$ one last time.

Every layer turns back into $\sqrt{x}$, so the whole thing equals $\sqrt{x}$.

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

The key is that $\sqrt[3]{x \cdot \sqrt{x}}$ gives back $\sqrt{x}$: the square root of $x$ feeds into the step and comes out unchanged. Once one layer produces $\sqrt{x}$, every layer above it does too. When a problem repeats the same step many times, work out one step and look for something that stays the same.

Test a number. Take $x = 64$, so $\sqrt{64} = 8$. The innermost piece is $\sqrt[3]{64 \cdot 8} = \sqrt[3]{512} = 8$. The next layer is again $\sqrt[3]{64 \cdot 8} = 8$, and so is the outer one. The answer for $x = 64$ is $8 = \sqrt{64}$. Choice (B) would give $16$, and choices (C), (D) and (E) are not $8$ either, so only (A) matches.

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