How do you factor #729x^6 - 4096y^6#? Algebra Polynomials and Factoring Factor Polynomials Using Special Products 1 Answer A. S. Adikesavan May 9, 2016 #(3x-4y)(3x+4y)(9x^2+12xy+16y^2)# Explanation: Use #(a^3-b^3)=(a-b)(a^2+ab+c^2)# Here, #729x^6-4096y^6# #=(3x)^6-(4y)^6# #=(a^3-b^3)#, where #a=(3x)^2 and b=(4y)^2# #=(a-b)(a^2+ab+b^2)# Now, use #(a^2-b^2)=(a-b)(a+b)# So, #729x^6-4096y^6=(3x-4y)(3x+4y)(9x^2+12xy+16y^2)# 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 1580 views around the world You can reuse this answer Creative Commons License