How do you simplify #7/(""^3sqrt(5)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer MAXIMILIAN C. May 9, 2018 #=(7*"3rd root"25)/5# Explanation: #7/("3rd root"5)# Multiply be #1/1# as #("3rd root"25)/("3rd root"25)# #=(7*"3rd root"25)/("3rd root"125)# #=(7*"3rd root"25)/5# Sorry, I don't know how to use 3rd root symbols. Answer link Related questions How do you simplify #\frac{2}{\sqrt{3}}#? How do you multiply and divide radicals? How do you rationalize the denominator? What is Multiplication and Division of Radicals? How do you multiply #(sqrt(a) +sqrt(b))(sqrt(a)-sqrt(b))#? How do you rationalize the denominator for #\frac{2x}{\sqrt{5}x}#? Do you always have to rationalize the denominator? How do you simplify #sqrt(5)sqrt(15)#? How do you simplify #(7sqrt(13) + 2sqrt(6))(2sqrt(3)+3sqrt(6))#? How do you simplify #(sqrt(7) + sqrt(5))^4 + (sqrt(7) - sqrt(5))^4#? See all questions in Multiplication and Division of Radicals Impact of this question 15044 views around the world You can reuse this answer Creative Commons License