How do you factor #x^6(y^2-1) - x^3(y^2-1)#? Algebra Polynomials and Factoring Factoring Completely 1 Answer Konstantinos Michailidis Sep 14, 2015 Refer to exlpanation Explanation: It is #x^6(y^2-1)-x^3(y^2-1)=x^3(y^2-1)(x^3-1)=x^3(y-1)(y+1)(x-1)(x^2+x+1)# Answer link Related questions What is Factoring Completely? How do you know when you have completely factored a polynomial? Which methods of factoring do you use to factor completely? How do you factor completely #2x^2-8#? Which method do you use to factor #3x(x-1)+4(x-1) #? What are the factors of #12x^3+12x^2+3x#? How do you find the two numbers by using the factoring method, if one number is seven more than... How do you factor #12c^2-75# completely? How do you factor #x^6-26x^3-27#? How do you factor #100x^2+180x+81#? See all questions in Factoring Completely Impact of this question 1014 views around the world You can reuse this answer Creative Commons License