Answer to Question #144625 in Algebra for Saaketh

Question #144625

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

Expert's answer

"\\begin{aligned}\n\\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)}\n\\\\&=\\frac{16x^9 y^{-4}}{48x^3y^3} = \\frac{x^6}{3y^7}\n\\end{aligned}"

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