On factoring $2+128(a+b)^3$ Factor $2+128(a+b)^3$
$=2[1+64(a+b)^3]= 2[1+64(a^3+3a^2b+3ab^2+b^3)]$
Then I'm at a loss for what to do.
$=2[1+64(a^3+b^3)+192ab(a+b)]= 2[1+64(a+b)(a^2-ab+b^2)+192ab(a+b)]=2[1+64(a+b)(a^2-ab+b^2+3ab)]=2[1+64(a+b)^3]$
Back to where I began...
But the answer is given as $2[1+4a+4b][1-4(a+b)+16(a+b)^2]$
I cannot comprehend how it's possible to factor into this answer. Please help me. Thank you.

You're almost there, in fact consider
$$2[1+64(a+b)^3] $$
and recall that  $x^3+y^3=(x+y)(x^2-xy+y^2)$. In your case, take $x=1$ and $y=4(a+b)$.
Hence you have:
$$2[1+64(a+b)^3]=2[1+4(a+b)][1-4(a+b)+16(a+b)^2]  $$
