How do you simplify #(3x+4)^3#? Algebra Polynomials and Factoring Multiplication of Polynomials by Binomials 1 Answer Gerardina C. Oct 9, 2017 #=27x^3+108x^2+144x+64# Explanation: Since #(A+B)^3=A^3+3A^2B+3AB^2+B^3# we get #(3x+4)^3=(3x)^3+3(3x)^2*4+3(3x)*4^2+4^3=# #=27x^3+108x^2+144x+64# Answer link Related questions What is FOIL? How do you use the distributive property when you multiply polynomials? How do you multiply #(x-2)(x+3)#? How do you simplify #(-4xy)(2x^4 yz^3 -y^4 z^9)#? How do you multiply #(3m+1)(m-4)(m+5)#? How do you find the volume of a prism if the width is x, height is #2x-1# and the length if #3x+4#? How do you multiply #(a^2+2)(3a^2-4)#? How do you simplify #(x – 8)(x + 5)#? How do you simplify #(p-1)^2#? How do you simplify #(3x+2y)^2#? See all questions in Multiplication of Polynomials by Binomials Impact of this question 3656 views around the world You can reuse this answer Creative Commons License