Infinite Nested Radicals

$ a) $ $$ \bullet \quad \large \sqrt[n]{ a\sqrt[n]{ a\sqrt[n]{ a\sqrt[n]{ a \cdots } } } } = \sqrt[n-1]{ a} $$

$ \large \sqrt[n]{a \underbrace{\sqrt[n]{a \sqrt[n]{a \sqrt[n]{a \cdots}}}}_{x}} = \sqrt[n-1]{a} \Rightarrow \sqrt[n]{ a \cdot x } = x $

$ \large x = \sqrt[n-1]{a } $

The validity of this can be demonstrated as shown above.

Examples:

$ \bullet \quad \sqrt[4]{ 8\sqrt[4]{ 8\sqrt[4]{ 8} \cdots }} = \sqrt[4-1]{ 8} = 2 $

$ \bullet \quad \sqrt{ 7\sqrt{ 7\sqrt{ 7} \cdots }} = \sqrt[2-1]{ 7} = 7 $

Question 1

Evaluate the expression: $$ \sqrt[4]{ 3\sqrt{ 3\sqrt[4]{ 8 \sqrt{3 \cdots } } }} $$

\[
\text{A) } \sqrt[7]{ 27} \quad
\text{B) } \sqrt[7]{ 9} \quad
\text{C) } \sqrt[7]{ 3} \quad
\text{D) } 3 \quad
\text{E) } 1
\]

Solution

$$ \sqrt[4]{ 3\sqrt{ 3\sqrt[4]{ 8 \sqrt{3 \cdots } } }} = \sqrt[4]{ \sqrt{3 \cdot 3^2 \sqrt[4]{ \sqrt{ 3 \cdot 3^2} \cdots } } }=\sqrt[8]{27 \sqrt[8]{27 \cdots } } $$

$$ \sqrt[8-1]{27 } = \sqrt[7]{