How do you simplify #root5(2)/(3root5(162))#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Nityananda Jan 16, 2017 #3^-(9/5)# Explanation: Given, #root(5)2/[3xx root(5) 162]# #rArr 2^(1/5)/[3xx(162)^(1/5)]# #rArr 2^(1/5)/[3xx (2xx81)^(1/5)# #rArr 2^(1/5)/[3xx2^(1/5)xx(3^4)^(1/5)# #rArr cancel[2^(1/5)]/[cancel[2^(1/5)] xx3^1 xx 3^(4/5)# #rArr 1/[3^(1+4/5)# #rArr 1/[3^{(5+4)/5}]# #rArr 1/3^(9/5)# #rArr 3^-(9/5)# 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 simplify #7/(""^3sqrt(5)#? 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))#? See all questions in Multiplication and Division of Radicals Impact of this question 1088 views around the world You can reuse this answer Creative Commons License