Answer
$100(m+1)(m-1)(m^2-m+1)(m^2+m+1)$
Work Step by Step
Factoring the $GCF=
100
,$ the given $\text{
expression,
}$ $
100m^6-100
,$ is equivalent to
\begin{align*}
100(m^6-1)
.\end{align*}
Using $a^2-b^2=(a+b)(a-b),$ the $\text{
expression
}$ above is equivalent to
\begin{align*}
&
100[(m^3)^2-(1)^2]
\\&=
100[(m^3+1)(m^3-1)]
.\end{align*}
The expressions $
m^3
$ and $
1
$ are both perfect cubes (the cube root is exact). Hence, $
(m^3+1)
$ is a $\text{
sum
}$ of $2$ cubes.
The expressions $
m^3
$ and $
1
$ are both perfect cubes (the cube root is exact). Hence, $
(m^3-1)
$ is a $\text{
difference
}$ of $2$ cubes.
\begin{align*}
a^3+b^3&=(a+b)(a^2-ab+b^2)
\\&\text{ or }\\
a^3-b^3&=(a-b)(a^2+ab+b^2)
,\end{align*}
The expression above is equivalent to
\begin{align*}
&
100[(m)^3+(1)^3][(m)^3-(1)^3]
\\&=
100[(m+1)(m^2-m+1)][(m-1)(m^2+m+1)]
\\&=
100(m+1)(m-1)(m^2-m+1)(m^2+m+1)
.\end{align*}
Hence, the factored form of $100m^6-100$ is $
100(m+1)(m-1)(m^2-m+1)(m^2+m+1)
$.
