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?

This is correct but you are probably asked to continue with the identity 
$$
x^4-4x^2+16=x^4+8x^2+16-12x^2=(x^2+4)^2-12x^2=a^2-b^2,
$$
for some $a$ and $b$ I will let you discover. The final factorisation of $x^6+64$ over the field $\mathbb R$ is the product of three polynomials $x^2+px+q$ with $p^2\lt4q$. 
Recall that, together with the polynomials of degree $1$, these are the only irreducible polynomials over $\mathbb R$, hence every real polynomial is a multiple of the product of some degree $2$ polynomials $x^2+px+q$ with $p^2\lt4q$ and some degree $1$ polynomials $x-c$.
