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: 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 $x-c$.
A: You are fine.  You can go further with the $x^4-4x^2+16$ term, but it isn't clean or easy to find.
A: 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 $\bf R$.
