Simplifying $\sqrt{a^{4 n}} \cdot \frac{3}{a^{n}} \cdot a^{1+n} \cdot(2 a)^{3 n-1}$ as far as possible, using only positive integers So far here is my steps but I am unsure of how to do the final step
$$\sqrt{a^{4 n}} \cdot \frac{3}{a^{n}} \cdot a^{1+n} \cdot(2 a)^{3 n-1}$$
$$\left(a^{4 n}\right)^{\frac{1}{2}} \cdot \frac{3 a}{a^{n}} \cdot a^{1+n} \cdot a \cdot \frac{(2 a)^{3 n}}{2 a}
$$
by canceling the a's we therefore get
$$ \frac{a^{2 n} \cdot 3 \cdot 2^{3 n} \cdot(a)^{3 n}}{2} $$
How would you then express this in all positive integers

Just to clarify the comments, I have a full worked solution provided.
\begin{align}
&\quad\sqrt{a^{4n}} \cdot \frac{3}{a^{n}} \cdot a^{1+n} \cdot(2 a)^{3 n-1}\\
&=\left(a^{4n}\right)^{\frac{1}{2}}\cdot3\cdot\frac{1}{a^n}\cdot a^{1+n}\cdot2^{3n-1}\cdot a^{3n-1}\\
&=a^{\frac{4n}{2}}\cdot3\cdot a^{-n}\cdot a^{n+1}\cdot2^{3n-1}\cdot a^{3n-1}\\
&=3\cdot a^{2n}\cdot a^{-n}\cdot a^{n+1}\cdot2^{3n-1}\cdot a^{3n-1}\\
&=3\cdot2^{3n-1}\cdot a^{2n-n+(n+1)+(3n-1)}\\
&=3\cdot2^{3n-1}\cdot a^{2n-n+n+1+3n-1}\\
&=3\cdot2^{3n-1}\cdot a^{5n}\\
&=\frac{3}{2}a^{5n}\cdot2^{3n}
\end{align}
