How do you simplify #\frac { ( 4p ^ { 5} r ^ { 3} ) ^ { 2} } { ( 2p ^ { 4} ) ( 3p r ^ { 6} ) }#? Prealgebra Exponents, Radicals and Scientific Notation Exponents 1 Answer Shwetank Mauria Jul 15, 2017 #(4p^5r^3)^2/((2p^4)(3pr^6))=(8p^5)/3# Explanation: #(4p^5r^3)^2/((2p^4)(3pr^6))# = #(4^2(p^5)^2(r^3)^2)/(2xx3xx(p^4xxp)xxr^6)# = #(16p^(5xx2)r^(3xx2))/(6p^(4+1)r^6)# = #(16p^10r^6)/(6p^5r^6)# = #(2xx8p^10r^6)/(2xx3p^5r^6)# = #(cancel2xx8p^10cancel(r^6))/(cancel2xx3p^5cancel(r^6))# = #(8p^(10-5))/3# = #(8p^5)/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 1503 views around the world You can reuse this answer Creative Commons License