Given that
[tex]\begin{gathered} (\frac{5n^6}{m^{-3}n^2})^2 \\ \text{According to the law of indies,} \\ (\frac{a}{b})^2\text{ = }\frac{(a)^2}{(b)^2} \\ (\frac{5n^6}{m^{-3}n^2})\text{ = }\frac{(5n^6)^2}{(m^{-3}n^2)^2} \\ \text{According to the law of indicies again,} \\ (a^x)^y=a^{x\cdot y} \\ =\text{ }\frac{(5n^{6\text{ x 2}})}{(m^{-3\cdot2}n^{2\cdot2})} \\ =\text{ }\frac{5n^{12}}{(m^{-6}n^4)} \\ \text{According to the law of indices} \\ x^{-1}\text{ = }\frac{1}{x} \\ =5^{}n^{12}\cdot n^{-4}\cdot m^6 \\ =5n^{12\text{ - 4}}m^6 \\ =5n^8m^6 \end{gathered}[/tex]