How do you factor #a^3b^3 - 8x^6y^9#? Algebra Polynomials and Factoring Factor Polynomials Using Special Products 1 Answer George C. May 28, 2015 This is a difference of cubes: #a^3b^3 - 8x^6y^9# #=(ab)^3 - (2x^2y^3)^3# #=((ab)-(2x^2y^3))((ab)^2+(ab)(2x^2y^3)+(2x^2y^3)^2)# #=(ab-2x^2y^3)(a^2b^2+2abx^2y^3+4x^4y^6)# using the identity #A^3-B^3 = (A-B)(A^2+AB+B^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 1315 views around the world You can reuse this answer Creative Commons License