What is the simplest radical form of #sqrt216#? Algebra Radicals and Geometry Connections Simplification of Radical Expressions 2 Answers GiĆ³ Apr 20, 2015 #sqrt(216)=sqrt(4*54)=2sqrt(54)=2sqrt(6*9)=2*3sqrt(6)=6sqrt(6)# Answer link Meave60 Apr 20, 2015 #6 xx 6 xx 6=216# #sqrt216# = #sqrt(6 xx 6 xx 6)# = #sqrt(6xx6)sqrt6# = #6sqrt6# Answer link Related questions How do you simplify radical expressions? How do you simplify radical expressions with fractions? How do you simplify radical expressions with variables? What are radical expressions? How do you simplify #root{3}{-125}#? How do you write # ""^4sqrt(zw)# as a rational exponent? How do you simplify # ""^5sqrt(96)# How do you write # ""^9sqrt(y^3)# as a rational exponent? How do you simplify #sqrt(75a^12b^3c^5)#? How do you simplify #sqrt(50)-sqrt(2)#? See all questions in Simplification of Radical Expressions Impact of this question 1827 views around the world You can reuse this answer Creative Commons License