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 $(1-2m+4m^2)(1+2m+4m^2)$?

There is a simpler approach.  We could first factor $1 - 64m^6$ as a difference of squares, then factoring the resulting sum of cubes and difference of cubes.
\begin{align*}
1 - 64m^6 & = (1 + 8m^3)(1 - 8m^3)\\
          & = (1 + 2m)(1 - 2m + 4m^2)(1 - 2m)(1 + 2m + 4m^2)
\end{align*}
As for your method of first factoring $1 - 64m^3$ as a difference of cubes, we can add and subtract $4m^2$ to the term $1 + 4m^2 + 16m^4$ to form a difference of squares.
\begin{align*}
1 - 64m^6 & = (1 - 4m^2)(1 + 4m^2 + 16m^4)\\
          & = (1 - 4m^2)(1 + 8m^2 + 16m^4 - 4m^2)\\
          & = (1 - 4m^2)[(1 + 4m^2)^2 - 4m^2]\\
          & = (1 + 2m)(1 - 2m)(1 + 4m^2 + 2m)(1 + 4m^2 - 2m)\\
          & = (1 + 2m)(1 - 2m)(1 + 2m + 4m^2)(1 - 2m + 4m^2)
\end{align*}
