[tex]\bf \sqrt{48n^9}\qquad\begin{cases}
48=2\cdot 2\cdot 2\cdot 2\cdot 3\\
\qquad 2^2\cdot 2^2\cdot 3\\
\qquad (2^2)^2\cdot 3\\
n^9=n^{8+1}\\
\qquad n^8\cdot n^1\\
\qquad n^{4\cdot 2}\cdot n\\
\qquad (n^4)^2\cdot n
\end{cases}\implies \sqrt{(2^2)^2\cdot 3\cdot (n^4)^2\cdot n}
\\\\\\
2^2n^4\sqrt{3n}\implies 4n^4\sqrt{3n}[/tex]
