If N>1, then √[3]N√[3]N√[3]N=

1998 AMC 12 Problem 7

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N\gt1,

then

\sqrt[3]{N\sqrt[3]{N\sqrt[3]{N}}}=

N^{\frac{1}{27}}

N^{\frac{1}{9}}

N^{\frac{1}{3}}

N^{\frac{13}{27}}

N

Start with the innermost cube root and convert it to an exponent

At each outer level, add

1

to the exponent and divide by

3

The innermost radical is

N^{\frac{1}{3}}.

The next is

(N\cdot N^{\frac{1}{3}})^{\frac{1}{3}}=N^{\frac{4}{9}}.

The outermost is

(N\cdot N^{\frac{4}{9}})^{\frac{1}{3}}=N^{\frac{13}{27}}.

Thus D is correct.