How do you factor #x^15 - 1/64#? Algebra Polynomials and Factoring Factor Polynomials Using Special Products 1 Answer Shwetank Mauria Apr 28, 2016 #x^15-1/64=(x^5-1/4)(x^(10)+1/4x^5+1/16)# Explanation: #x^15-1/64=(x^5)^3-(1/4)^3# and hence using identity #a^3-b^3=(a-b)(a^2+ab+b^2)#, we get #(x^5)^3-(1/4)^3=(x^5-1/4)((x^5)^2+1/4x^5+(1/4)^2)# = #(x^5-1/4)(x^(10)+1/4x^5+1/16)# 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 1621 views around the world You can reuse this answer Creative Commons License