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)

There are two errors in the factorization
$$x^6-64=(x^2-4)(x^4+4x-16)$$
The constant term in the quartic should be $+16$ instead of $-16$, and the linear term, $4x$, should be a quadratic term, $4x^2$.  That is, the correct factorization is
$$x^6-64=(x^2-4)(x^4+4x^2+16)$$
This now agrees with the factorization $(x+2)(x^2-2x+4)(x-2)(x^2+2x+4)$ since $(x-2)(x+2)=x^2-4$ and
$$\begin{align}
(x^2-2x+4)(x^2+2x+4)&=(x^2+4-2x)(x^2+4+2x)\\
&=(x^2+4)^2-(2x)^2\\
&=x^4+8x^2+16-4x^2\\
&=x^4+4x^2+16
\end{align}$$
