Question

Simplify $\sqrt[3]{x\sqrt[3]{x\sqrt[3]{x\sqrt{x}}}}.$

$\sqrt{x}$

$\sqrt[3]{x^2}$

$\sqrt[27]{x^2}$

$\sqrt[54]{x}$

$\sqrt[81]{x^{80}}$

Rewrite each radical using fractional exponents, working from the inside out

The innermost $x\sqrt{x} = x^{\frac{3}{2}},$ and each cube root divides the exponent by $3$

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.