How do you factor #27x^3 - 8y^3 #? Algebra Polynomials and Factoring Factor Polynomials Using Special Products 1 Answer Meave60 Jan 15, 2016 #27x^3-8y^3=(3x)^3-(2y)^3=color(red)((3x-2y))color(red)((9x^2+6xy+4y^2))# Explanation: #27x^3-8y^3# fits the form of the difference of cubes, #a^3-b^3#, where #a=3x# and #b=2y#. #a^3-b^3=(3x)^3-(2y)^3# #a^3-b^3=(a-b)(a^2+ab+b^2)# Substitute the known values for #a# and #b#. #(3x)^3-(2y)^3=(3x-2y)((3x)^2+(3x*2y)+(2y)^2)# Simplify. #(3x)^3-(2y)^3=(3x-2y)(9x^2+6xy+4y^2)# Answer link Related questions How do you factor special products of polynomials? How do you identify special products when factoring? How do you factor #x^3 -8#? What are the factors of #x^3y^6 – 64#? How do you know if #x^2 + 10x + 25# is a perfect square? How do you write #16x^2 – 48x + 36# as a perfect square trinomial? What is the difference of two squares method of factoring? How do you factor #16x^2-36# using the difference of squares? How do you factor #2x^4y^2-32#? How do you factor #x^2 - 27#? See all questions in Factor Polynomials Using Special Products Impact of this question 1320 views around the world You can reuse this answer Creative Commons License