Factor $1-64m^6$ under $\mathbb{Q}$

Question

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

Answer 1

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*}

Answer 2

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