Factoring the expression $x^6 + 64$ Alright, so apparently I've factored this out wrong...
$x^6 + 64 =$
$x^6 + 2^6$
Then I continued, using $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ to get ...
$$(x^2)^3 + 64 =
(x^2)^3 + 4^3 =
(x^2 + 4)(x^4 - 4x^2 + 16)$$
How is this incorrect?

A different approach is to first factorize it over complex numbers:
$$x^6+2^6=\prod_{k=0}^5(x-2\cdot e^{\pi i(1/6+k/3)}) $$
And then pair conjugate roots, yielding:
$$\prod_{k=0}^2(x-2\cdot e^{\pi i(1/6+k/3)})(x-2\cdot e^{-\pi i(1/6+k/3)})=\prod_{k=0}^2(x^2-4\cos(\pi(1/6+k/3))+4)$$
Which equals
$$(x^2-2\sqrt 3x+4)(x^2+4)(x^2+2\sqrt 3x+4)$$
This method will work for any polynomial over $\bf R$ you can factor over $\bf C$ into linear factors (such a factorization always exists, but sometimes it might be impossible to express it in the way you would like).
The reason it will work is that for any real polynomial, it's non-real complex roots occur always occur in conjugate pairs, and for a nonreal complex $z$ we have that $(x-z)(x-\overline z)=x-2\Re(z)+\lvert z\rvert^2$ and the latter is indecomposable over $\bf R$.
