How do you simplify #sqrt(44/9)*sqrt(18/7)*sqrt(35/72)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Shwetank Mauria Feb 18, 2017 #sqrt(44/9)*sqrt(18/7)*sqrt(35/72)=1/3sqrt55# Explanation: #sqrt(44/9)*sqrt(18/7)*sqrt(35/72)# = #sqrt((2xx2xx11)/(3xx3))*sqrt((2xx3xx3)/7)*sqrt((7xx5)/(2xx2xx2xx3xx3))# = #sqrt((cancel(2xx2)xx11)/(cancel(3xx3)))*sqrt((cancel2xxcancel(3xx3))/cancel7)*sqrt((cancel7xx5)/(cancel2xxcancel(2xx2)xx3xx3))# = #sqrt((11xx5)/(3xx3)# = #1/3sqrt55# 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 1108 views around the world You can reuse this answer Creative Commons License