How do you multiply #(x^2-1)^3#? Algebra Polynomials and Factoring Multiplication of Polynomials by Binomials 1 Answer kamil9234 May 1, 2018 #(x^2-1)^3=x^6-3x^4+3x^2-1# Explanation: #(a-b)^3=a^3-3a^2b+3ab^2-b^3# #(x^2-1)^3=(x^2)^3-3(x^2)^2+3x^2-1=x^6-3x^4+3x^2-1# Answer link Related questions What is FOIL? How do you use the distributive property when you multiply polynomials? How do you multiply #(x-2)(x+3)#? How do you simplify #(-4xy)(2x^4 yz^3 -y^4 z^9)#? How do you multiply #(3m+1)(m-4)(m+5)#? How do you find the volume of a prism if the width is x, height is #2x-1# and the length if #3x+4#? How do you multiply #(a^2+2)(3a^2-4)#? How do you simplify #(x – 8)(x + 5)#? How do you simplify #(p-1)^2#? How do you simplify #(3x+2y)^2#? See all questions in Multiplication of Polynomials by Binomials Impact of this question 14368 views around the world You can reuse this answer Creative Commons License