How do you simplify #sqrt8/sqrt6#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Shwetank Mauria Jun 16, 2016 #sqrt8/sqrt6=(2sqrt3)/3# Explanation: #sqrt8/sqrt6=sqrt((2xx2xx2)/(2xx3))# = #sqrt((2xx2xxcancel2)/(cancel2xx3))# = #2/sqrt3xxsqrt3/sqrt3# = #(2sqrt3)/3# 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 1281 views around the world You can reuse this answer Creative Commons License