Complete factorization of $64y^6 - 1$ If we use the difference of two cubes identity, then we have
$$64y^6 - 1 = (4y^2-1)(16y^4+4y^2+1) = (2y+1)(2y-1)(16y^4+4y^2+1)$$ which is the suggested solution of the textbook.
But if we use the difference of two squares identity, we then have
$$64y^6 - 1 = (8y^3+1)(8y^3-1) = (2y+1)(2y-1)(4y^2-2y+1)(4y^2+2y+1).$$
I then came up with the question: what is complete factorization? Of course we can do this trick:
$$16y^4+4y^2+1 = 16y^4+8y^2+1 - 4y^2 = (4y^2+1 - 2y)(4y^2+1+2y).$$
How do we know when to stop factorizing?

The answer   depends on the base ring/field you consider:
The complete factorisation over $\mathbf Z, \mathbf Q$ or $\mathbf R$ is indeed
$$(2x-1)(2x+1)(4x^2+2x+1)(4x^2-2x+1)$$
But on $\mathbf C$, you obtain a product of linear factors involving the sixth roots of unity:
\begin{align}
64x^6-1&=\prod_{k=1}^6\Bigl(2x-\mathrm e^{\tfrac{2ik\pi}{6}}\Bigr)\\
&=\underbrace{\Bigl(2x-\mathrm e^{\tfrac{i\pi}{3}}\Bigr)\Bigl(2x-\mathrm e^{\tfrac{5i\pi}{3}}\Bigr)}_{\textstyle4x^2+2x+1}\,\underbrace{\Bigl(2x-\mathrm e^{\tfrac{2i\pi}{3}}\Bigr)\Bigl(2x-\mathrm e^{\tfrac{4i\pi}{3}}\Bigr)}_{\textstyle4x^2-2x+1}\,\Bigl(\underbrace{2x-\mathrm e^{\tfrac{3i\pi}{3}}\rule[-1.5ex]{0pt}{1.5ex}}_{\textstyle2x+1}\Bigr)\,\Bigl(\underbrace{2x-\mathrm e^{\tfrac{6i\pi}{3}}\rule[-1.5ex]{0pt}{1.5ex}}_{\textstyle2x-1}\Bigr)
\end{align}\,
