Question

factor: w^6-729

Answer 1

If you look at this as the difference of two squares you would get.
(w^3 + 27)(w3 - 27)
Now you have the sum and the difference of two cubes
There are formulas for this.
(a-b)^3 = (a-b)(a^2 + ab + b^2)
(a+b)^3 = (a+b)(a^2 - ab + b^2)
so in this case replace a with "w" and b with "3"
(w^3 + 27) = (w + 3)(w^2 - 3w + 9)
(w^3 - 27) = (w-3)(w^2 + 3w + 9)

so the final answer is the product of those two answers
(w + 3)(w - 3)(w^2 - 3w + 9)(w^2 + 3w + 9)