How do you simplify #(3x^-1y^3)^2 / ((3xy^-1)^3 (9x^-9y^5))#? Algebra Exponents and Exponential Functions Exponential Properties Involving Quotients 1 Answer Ananda Dasgupta Mar 22, 2018 #(x^4y^4)/27# Explanation: #(3x^-1y^3)^2 / ((3xy^-1)^3 (9x^-9y^5)) = (3^2(x^-1)^2(y^3)^2)/((3^3x^3(y^-1)^3)(9x^-9y^5))# #qquad =(3^2x^-2y^6)/((3^3x^3y^-3)(3^2x^-9y^5)) = (3^2x^-2y^6)/(3^(3+2)x^(3-9)y^(-3+5)) =(3^2x^-2y^6)/(3^5x^-6y^2) = 3^(2-5) x^(-2-(-6)) y^(6-2) = 3^-3x^4y^4=(x^4y^4)/27# Answer link Related questions What is the quotient of powers property? How do you simplify expressions using the quotient rule? What is the power of a quotient property? How do you evaluate the expression #(2^2/3^3)^3#? How do you simplify the expression #\frac{a^5b^4}{a^3b^2}#? How do you simplify #((a^3b^4)/(a^2b))^3# using the exponential properties? How do you simplify #\frac{(3ab)^2(4a^3b^4)^3}{(6a^2b)^4}#? Which exponential property do you use first to simplify #\frac{(2a^2bc^2)(6abc^3)}{4ab^2c}#? How do you simplify #(x^5y^8)/(x^4y^2)#? How do you simplify #[(2^3 *-3^2) / (2^4 * 3^-2)]^2#? See all questions in Exponential Properties Involving Quotients Impact of this question 1486 views around the world You can reuse this answer Creative Commons License