Factor $1-64m^6$ under $\mathbb{Q}$ Factor $1-64m^6$ under $\mathbb{Q}$. We can employ the well known formula: $a^3-b^3 = (a-b)\cdot(a^2+ab+b^2)$.
$1-64m^6 = (1)^3-(4m^2)^3 = (1-4m^2)(1+4m^2+16m^4)=(1+2m)(1-2m)(1+4m^2+16m^4)$.
However, the answer is provided as $(1+2m)(1-2m)(1-2m+4m^2)\cdot(1+2m+4m^2)$.
I cannot understand how it is possible to simplify $(1+4m^2+16m^4)$ to $(1-2m+4m^2)(1+2m+4m^2)$?

$$1+4m^2+16m^4$$
$$=1+2\cdot 4m^2+16m^4-4m^2$$
$$=(1+4m^2)^2-(2m)^2$$
$$=(1-2m+4m^2)(1+2m+4m^2)$$
