Question 766752
'(
x^9 + 1
Can be factored as the "sum of cubes" which is x^3 + y^3 = (x + y)(x^2- xy + y^2)
so we have
x^9 + 1 = (x^3 + 1)(x^6 - x^3 + 1)
:
2x^9 + 16
Factor out 2, then factor as the sum of cubes
2(x^9 + 8) = 2(x^3 + 2)(x^6 - 2x^3 + 4)