Problem: If N>1N>1N>1, then NNN333=\sqrt[3]{N \sqrt[3]{N \sqrt[3]{N}}}=3N3N3N=
Answer Choices:
A. N127N^{\frac{1}{27}}N271
B. N19N^{\frac{1}{9}}N91
C. N13N^{\frac{1}{3}}N31
D. N1327N^{\frac{13}{27}}N2713
E. NNN
Solution:
NNN333=NN⋅N1333=NN4333=N⋅N193=N1393=N1327\sqrt[3]{N \sqrt[3]{N \sqrt[3]{N}}}=\sqrt[3]{N \sqrt[3]{N \cdot N^{\frac{1}{3}}}}=\sqrt[3]{N \sqrt[3]{N^{\frac{4}{3}}}}=\sqrt[3]{N \cdot N^{\frac{1}{9}}}=\sqrt[3]{N^{\frac{13}{9}}}=N^{\frac{13}{27}} 3N3N3N=3N3N⋅N31=3N3N34=3N⋅N91=3N913=N2713
OR\textbf{OR} OR
NNN333=(N(N(N)13)13)13=(N(N13⋅N19))13=N13⋅N13⋅N127=N11327\sqrt[3]{N \sqrt[3]{N \sqrt[3]{N}}}=\left(N\left(N(N)^{\frac{1}{3}}\right)^{\dfrac{1}{3}}\right)^{\frac{1}{3}}=\left(N\left(N^{\frac{1}{3}} \cdot N^{\frac{1}{9}}\right)\right)^{\frac{1}{3}}=N^{\frac{1}{3}} \cdot N^{\frac{1}{3}} \cdot N^{\frac{1}{27}}=N^{\frac{113}{27}} 3N3N3N=⎝⎜⎜⎛N(N(N)31)31⎠⎟⎟⎞31=(N(N31⋅N91))31=N31⋅N31⋅N271=N27113