What is the real cube root of $3^{3^{333}}$?

Question

What is the real cube root of $\displaystyle 3^{3^{333}}$?

The answer says it's $\displaystyle 3^{3^{332}}$, but I don't know how to get the answer myself.

What are the steps?

Thank you.

Answer 1

Just do a simple substitution. Let $u=3^{333}$. This means

$$\sqrt[3]{3^{3^{333}}}=\sqrt[3]{3^u}=3^{\frac{u}{3}}$$

where

$$\frac{u}{3}=\frac{3^{333}}{3}=3^{332}$$

Answer 2

The cubed root of $x$ is $x^{1/3}$.

You will also need the rule $(a^b)^c = a^{bc}$ and also note that $\frac{1}{3}3^n = 3^{n-1}$.

The cubed root of $3^{3^{333}}$ is $(3^{3^{333}})^{1/3} = 3^{(\frac{1}{3}3^{333})} = 3^{3^{333-1}} = 3^{3^{332}}$.

Answer 3

\begin{align*} \sqrt[3]{3^{3^{333}}} &= 3^{\frac{1}{3}(3^{333})} \\ &= 3^{3^{-1}(3^{333})} \\ &= 3^{3^{-1 + 333}} \\ &= 3^{3^{332}} \text{.} \end{align*}