How do you factor #64-a^2b^2#? Algebra Polynomials and Factoring Factor Polynomials Using Special Products 1 Answer Alan P. Jul 7, 2015 #64-a^2b^2 = (8+ab)(8-ab)# Explanation: #64 - a^2b^2# #color(white)("XXXX")##=8^2 - (ab)^2# This is the difference of squares with factors: #color(white)("XXXX")##=(8+ab)(8-ab)# 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 1423 views around the world You can reuse this answer Creative Commons License