Factor x^9+512?

Question

x^9+512

I got it down to (x+2)(x^2-2x+4)(x^6-8x^3+64)...Is this completely factored? Or is it still factorable?

Answer 1

(x^3)^3 + (8)^3
(x^3 + 8)(x^6 - 8x^3 + 64)
(x)^3 + (2)^2
(x + 2)(x^2 - 2x + 4)(x^6 - 8x^3 + 64)
x^3 = u
(x + 2)(x^2 - 2x + 4)(u^2 - 8u + 64) No factoring here ...
(x + 2)(x^2 - 2x + 4)(x^6 - 8x^3 + 64) is factored

Answer 2

x^9 + 512 = (x^3)^3 + 8^3
= (x^3 + 8)((x^3)^2 - 8(x^3) + 8^2)
= (x^3 + 2^3)((x^3)^2 - 8(x^3) + 64)
= (x + 2)(x^2 - 2x + 4)((x^3)^2 - 8(x^3) + 64),
in x^2 - 2x + 4, (-2)^2 - 4(1)(4) = -12 < 0, so x^2 - 2x + 4 can't be factored further in R,
in (x^3)^2 - 8(x^3) + 64, (-8)^2 - 4(1)(64) = -192 < 0, so
(x^3)^2 - 8(x^3) + 64 can't be factored further in R,
thus x^9 + 512 = (x + 2)(x^2 - 2x + 4)((x^3)^2 - 8(x^3) + 64)
= (x + 2)(x^2 - 2x + 4)(x^6 - 8(x^3) + 64).

Answer 3

the determinant of the second and third terms are negative. That means they don't have real solutions.
Over complex numbers, of course is possible and easy.
the third one is factorable even if its roots are nonreal in the form degree 2+ degree 2+ degree2. You have to find the roots and then the polynomials of degree 2 are easy to find.
Over rationals though, the polynomials is not more factorable.

Answer 4

Completely factored, you can check with the quadratic equation (the b^2-4ac). When you do that if you get a negative answer then you know it is factored.