How do you simplify #(14x^2y^2z^2)/(21xy^8z^4)#? Algebra Exponents and Exponential Functions Exponential Properties Involving Quotients 1 Answer Alan P. Jun 15, 2015 #(14x^2y^2z^2)/(21xy^8z^4)= x/(7y^6z^2)# Explanation: #(14x^2y^2z^2)/(21xy^8z^4)# can be grouped as similar components: #color(white)("XXXX")##=(14/21) * ((x^2)/(x)) * ((y^2)/(y^8)) * ((z^2)/(z^4))# #color(white)("XXXX")##= 1/7 * x/1 * 1/(y^6) * 1/(z^2)# #color(white)("XXXX")##= x/(7y^6z^2)# 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 1716 views around the world You can reuse this answer Creative Commons License