How do you factor #x^6 - 27y^3 - x^4 - 3x^2y - 9y^2#? Algebra Polynomials and Factoring Factoring Completely 1 Answer George C. Jun 1, 2015 #x^6-27y^3-x^4-3x^2y-9^2# #=(x^2)^3-(x^2)^2-(x^2)(3y)-(3y)^2-(3y)^3# #= X^3-X^2-XY-Y^2-Y^3# ...where #X=x^2# and #Y=3y#. #= (X^3-Y^3)-(X^2+XY+Y^2)# #= (X-Y)(X^2+XY+Y^2) - (X^2+XY+Y^2)# #= (X-Y-1)(X^2+XY+Y^2)# #= ((x^2)-(3y)-1)((x^2)^2+(x^2)(3y)+(3y)^2)# #= (x^2-3y-1)(x^4+3x^2y+9y^2)# 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 1519 views around the world You can reuse this answer Creative Commons License