How do you factor #8x^3*y^6 + 27#? Algebra Polynomials and Factoring Factor Polynomials Using Special Products 1 Answer mason m Dec 13, 2015 #(2xy^2+3)(4x^2y^4-6xy^2+9)# Explanation: Sum of cubes: #(a^3+b^3)=(a+b)(a^2-ab+b^2)# Since #8x^3y^6+27=(2xy^2)^3+(3)^3#, #a=2xy^2# #b=3# #8x^3y^6+27=(2xy^2+3)(4x^2y^4-6xy^2+9)# 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 1700 views around the world You can reuse this answer Creative Commons License