Write with one radical $\sqrt[4]{2^6}\cdot\sqrt{3^3}$ 
Write with one root (radical) $$\sqrt[4]{2^6}\cdot\sqrt{3^3}$$

In this lesson we have learnt that when the roots exist then $$\sqrt[n]{a}=\sqrt[nk]{a^k}$$
Using that here, we have $\gcd(4,2)=2,$ so
$$\sqrt[4]{2^6}\cdot\sqrt[2\cdot2]{\left(3^{3}\right)^2}=\sqrt[4]{2^6}\cdot\sqrt[4]{3^6}=\sqrt[4]{2^6\cdot3^6}=\sqrt[4]{\left(2\cdot3\right)^6}=\sqrt[4]{6^6}$$
Their solution, though, goes as
$$\sqrt[4]{2^6}\cdot\sqrt{3^3}=\sqrt{2^3}\sqrt{3^3}=\sqrt{6^3}=6\sqrt6$$
They haven't calculated the $\gcd(4,2)$ of the indices $4$ and $2$ as in the previous examples (e.g. $\sqrt[4]{3}\cdot\sqrt[3]{2}=...=\sqrt[12]{432}$). What am I missing? Thank you!
 A: Both answers are mathematically equal.  What is different is just observation.

We see that,
$$\begin{align}\sqrt[4]{2^6}\cdot\sqrt{3^3}&=2^{\frac 64}\times 3^{\frac 32}\\
&=2^{\frac 32}\times 3^{\frac 32}\\
&=(2\times 3)^{\frac 32}\\
&=6^{\frac 32}\\
&=\sqrt{6^3}\\
&=\sqrt{6^2\times 6}\\
&=6\sqrt{6}\end{align}$$
Also, you can write
$$\begin{align}\sqrt[4]{2^6}\cdot\sqrt{3^3}&=2^{\frac 64}\times 3^{\frac 32}\\
&=2^{\frac 64}\times 3^{\frac 64}\\
&=(2\times 3)^{\frac 64}\\
&=6^{\frac 64}\\
&=\sqrt[4]{6^6}\end{align}$$
You already know that,
$$\begin{align}\sqrt[4]{6^6}&=6^{\frac 64}\\
&=6^{\frac 32}\\
&=\sqrt{6^3}\\
&=6\sqrt 6.\end{align}$$
A: The author of the solution notices that $6=2\cdot 3$, so $$\sqrt[4]{2^6}=\sqrt[4]{2^{2\cdot 3}}=\sqrt[4]{\left(2^3\right)^2}=\sqrt[4]{8^2}$$
By your provided equality, if we let $a=8, n=k=2$, then  we have
$$\sqrt{8}=\sqrt[4]{8^2}$$
Now, $8=2^3$, so we can conclude
$$\sqrt[4]{2^6}\sqrt{3^3}=\sqrt{2^3}\sqrt{3^3}$$

Also, we don't really need to write $2^{2\cdot 3}$ as $8^2$ to prove that $\sqrt[4]{2^6}=\sqrt{2^3}$. In fact, we have
$$\sqrt[nk]{a^{ck}}=\sqrt[n]{a^c}$$
This means we only need to compute the GCD of the radical and exponent, or in our case, $\gcd(4,6)$, to simplify radical expressions. I'll leave it to you to prove this yourself. (Hint: write $a^{ck}=\left(a^c\right)^k$ and let $b=a^c$)
