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?

Are you looking for the factorization in $\mathbb{R}[x]$?
Note that for all real values $y$,
$$4y^2+1 - 2y=3y^2+(y-1)^2>0\quad 4y^2+1+2y=3y^2+(y+1)^2>0$$
so they cannot be factored as $a(y-y_1)(y-y_2)$ with $a,y_1,y_2\in \mathbb{R}$, otherwise they should be zero for $y_1$ and $y_2$.
