Question

Simplify: [(6x^3 y^-4)^-2] / [(3x^2 y^5)^-3] (The / is the fraction line)

Answer 1

$\bf \cfrac{(6x^3y^{-4})^{-2}}{(3x^2y^5)^{-3}}\implies \stackrel{\textit{using positive exponents}}{\cfrac{(3x^2y^5)^3}{(6x^3y^{-4})^2}}\implies \stackrel{\textit{distributing the exponent}}{\cfrac{(3^3x^{2\cdot 3}y^{5\cdot 3})}{(6^2x^{3\cdot 2}y^{-4\cdot 2})}} \\\\\\ \cfrac{27x^6y^{15}}{36x^6y^{-8}}\implies \cfrac{27}{36}\cdot \cfrac{x^6y^{15}}{x^6y^{-8}}\implies \cfrac{3}{4}\cdot x^6x^{-6}y^{15}y^8\implies \cfrac{3}{4}\cdot x^{6-6}y^{15+8} \\\\\\ \cfrac{3}{4}\cdot 1y^{23}\implies \cfrac{3y^{23}}{4}$

Answer 2

Opening the brackets, we have:
6^-2 x^-6 y^8
__________
3^-3x^-6 y^-15

= -36 x^-6 y^ 8
__________
-27 x^-6 y^-15
=
27 x^6 y ^15
_________
36 x^6 y^-8

divide the coefficients by 9. Work out the variables.

= 3y^23
_____
4