Factoring the difference of two cubes?

Difference of 2 cubes: x^3 - y^3 = (x - y)(x^2 + xy + y^2) Let 64a^6 - 1 = x^3 - y^3 (i) x^3 = 64a^6, x = 4a^2 (ii) y^3 = 1, y = 1 With that 64a^6 - 1 = (4a^2 - 1)(4a^2)^2 + (4a^2)(1) + (1)^2 Now, 4a^2 - 1 is the difference of 2 squares, which can be broken into (2a + 1)(2a - 1). 64a^6 - 1 = (2a + 1)(2a - 1)(4a^2)^2 + (4a^2)(1) + (1)^2 Let's expand the terms in the square brackets 1st. 64a^6 - 1 = (2a + 1)(2a - 1)(16a^4 + 4a^2 + 1) Now, distribute (2a + 1) into (16a^4 + 4a^2 + 1).

64a^6 - 1 = (2a - 1)(2a + 1)*(16a^4 + 4a^2 + 1) You should get 64a^6 - 1 = (2a - 1)2a(16a^4 + 4a^2 + 1) + 1(16a^4 + 4a^2 + 1) 64a^6 - 1 = (2a - 1)(32a^5 + 8a^3 + 2a) + (16a^4 + 4a^2 + 1).

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