factoring $(x^6 - y^6)$: what is going on here? I apologize if this question already exists, but it was quite difficult to word.
In my math class today, we learned how to factor a difference of two perfect cubes. One of our practice questions was:
factor $x^6 - 64$
almost everyone other than me (including my teacher, as he was rushing) ended up with:
$(x^2 - 4)(x^4 + 4x^2 + 16)$
I pointed out two things:
1:
$(x^6 - 64)$ was also a difference of two perfect squares, and we could factor it into this:
$(x^3 + 8)(x^3 - 8)$
Which could then be factored further into:
$(x + 2)(x^2 - 2x + 4)(x - 2)(x^2 + 2x + 4)$
and 2:
the $(x^2 - 4)$ was also a difference of two perfect squares, and we could factor further: 
$(x + 2)(x - 2)(x^4 + 4x^2 + 16)$
Now for my question:

How are $(x + 2)(x^2 - 2x + 4)(x - 2)(x^2 + 2x + 4)$ and $(x + 2)(x - 2)(x^4 + 4x^2 + 16)$ equivalent?

The teacher didn't seem to know off the top of his head, and I can't figure it out after trying for a half an hour. I've heard that you can factor the sum of two perfect squares with imaginary numbers, so maybe i can do something there to help explain this?
edit: fixed instances of "- 16" to "+ 16", and "4x" to "4x^2" (thanks for pointing that out)

A complete factorization of $x^6 - y^6$ can be performed as follows:  $$\begin{align*}
x^6 - y^6 &= (x^3)^2 - (y^3)^2 \\
&= (x^3 - y^3)(x^3 + y^3) \\
&= (x-y)(x^2 + xy + y^2)(x+y)(x^2 - xy + y^2). \end{align*}$$
The quadratic factors are irreducible in the ring of polynomials with rational coefficients $\mathbb Q[x]$ (more informally, they don't factor "nicely").
We can recombine these factors in carefully chosen ways to get expressions that might have resulted from an incomplete factorization; e.g., if we let $a = x^2 + y^2$ and $b = xy$, then
$$\begin{align*} (x^2 + xy + y^2)(x^2 - xy + y^2) &= (a+b)(a-b) \\
&= a^2 - b^2 \\ 
&= (x^2+y^2)^2 - (xy)^2 \\
&= x^4 + 2x^2 y^2 + y^4 - x^2 y^2 \\
&= x^4 + x^2 y^2 + y^4. \end{align*}$$
And if we combine $(x-y)(x+y)$ we get $x^2 - y^2$; thus we recover the incomplete factorization of $x^6 - y^6$ as a difference of cubes $$(x^2)^3 - (y^2)^3 = (x^2 - y^2)(x^4 + x^2 y^2 + y^4).$$  There are of course other ways to combine these factors, each leading to a different expression, but all are algebraically equivalent.
