Question

[(4x^6)^3(4y^-8)/ (2x)^4(12y^3)^2]^1/2

Answer 1

$\begin{aligned} \left(\frac{(4x^6)^3(4y^{-8})}{(2x)^4(12y^3)^2}\right)^{\frac{1}{2}} &= \frac{(2x^3)^3(2y^{-4})}{(2x)^2(12y^3)} \\&=\frac{16x^9 y^{-4}}{48x^3y^3} = \frac{x^6}{3y^7} \end{aligned}$