How do you factor #13(x^6+ 1)^4#? Algebra Polynomials and Factoring Special Products of Polynomials 1 Answer MattyMatty Jan 24, 2018 #13 (x^2+1)^4 (x^4 - x^2 + 1)^4# Explanation: #a^3 + b^3 = (a+b)(a^2-ab+b^2)# #=> x^6 + 1 = (x^2 + 1)(x^4 - x^2 + 1)# #=> 13(x^6+1)^4 = 13 (x^2+1)^4 (x^4 - x^2 + 1)^4# Answer link Related questions What are the Special Products of Polynomials? What is a perfect square binomial and how do you find the product? How do you simplify by multiplying #(x+10)^2#? How do you use the special product for squaring binomials to multiply #(1/4t+2 )^2#? How do you use the special product of a sum and difference to multiply #(3x^2+2)(3x^2-2)#? How do you evaluate #56^2# using special products? How do you multiply #(3x-2y)^2#? How do you factor # -8x^2 +32#? How do you factor #x^3-8y^3#? How do you factor # x^3 - 1#? See all questions in Special Products of Polynomials Impact of this question 1806 views around the world You can reuse this answer Creative Commons License