How do you simplify #(6+sqrt 7)(6-sqrt 7)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Don't Memorise Feb 13, 2016 #=color(blue)(29# Explanation: #(6+sqrt7)(6-sqrt7)# The above expression is of the form #color(blue)((a+b)(a-b) = a^2 - b^2# #color(blue)(a=6, b=sqrt7# So, #(6+sqrt7)(6-sqrt7) = color(blue)(6^2 - (sqrt7)^2# #=36-7# #=29# 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 2263 views around the world You can reuse this answer Creative Commons License