How do you factor #y^3-64#? Algebra Polynomials and Factoring Factor Polynomials Using Special Products 1 Answer Meave60 Jan 30, 2016 #y^3-4^3=(y-4)(y^2+4y+16)# Explanation: #y^3-64# is a difference of cubes, #a^3– b^3 = (a – b)(a^2 + ab + b^2)#, where #a=y# and #b=4#. Rewrite the equation. #(y)^3-(4)^3=(y-4)(y^2+(y*4)+4^2)# Simplify. #y^3-4^3=(y-4)(y^2+4y+16)# 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 10782 views around the world You can reuse this answer Creative Commons License