How do you simplify #\frac { ( 2x y ^ { 4} ) ^ { 3} } { 2x y ^ { 5} \cdot y x ^ { 5} }#? Prealgebra Exponents, Radicals and Scientific Notation Exponents 1 Answer Shwetank Mauria Jul 24, 2017 #(2xy^4)^3/(2xy^5*yx^5)=(4y^6)/x^3# Explanation: #(2xy^4)^3/(2xy^5*yx^5)# = #(2^3xx x^3(y^4)^3)/(2x^(1+5)y^(5+1))# = #(8x^3y^12)/(2x^6y^6)# = #(cancel8^4y^(12-6))/(cancel2x^(6-3))# = #(4y^6)/x^3# Answer link Related questions How do you simplify #c^3v^9c^-1c^0#? How do you simplify #(- 1/5)^-2 + (-2)^-2#? How do you simplify #(4^6)^2 #? How do you simplify #3x^(2/3) y^(3/4) (2x^(5/3) y^(1/2))^3 #? How do you simplify #4^3ยท4^5#? How do you simplify #(5^-2)^-3#? How do you simplify and write #(-5.3)^0# with positive exponents? How do you factor #12j^2k - 36j^6k^6 + 12j^2#? How do you simplify the expression #2^5/(2^3 times 2^8)#? When can I add exponents? See all questions in Exponents Impact of this question 1427 views around the world You can reuse this answer Creative Commons License