Answer
$(m+1)(m-1)(m^2-m+1)(m^2+m+1)$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To factor the given expression, $
m^6-1
,$ use the factoring of the difference of $2$ squares. Then use the factoring of the sum/difference of $2$ cubes to factor further.
$\bf{\text{Solution Details:}}$
The expressions $
m^6
$ and $
1
$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $
m^6-1
,$ is a difference of $2$ squares. Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
(m^3)^2-(1)^2
\\\\=
(m^3+1)(m^3-1)
.\end{array}
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. Using the factoring of the sum or difference of $2$ cubes which is given by $a^3+b^3=(a+b)(a^2-ab+b^2)$ or by $a^3-b^3=(a-b)(a^2+ab+b^2)$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
[(m)^3+(1)^3](m^3-1)
\\\\=
[(m+1)(m^2-m+1)](m^3-1)
\\\\=
(m+1)(m^2-m+1)(m^3-1)
.\end{array}
The expressions $
m^3
$ and $
1
$ are both perfect cubes (the cube root is exact). Hence, $
m^311
$ is a $\text{
difference
}$ of $2$ cubes. Using the factoring of the sum or difference of $2$ cubes which is given by $a^3+b^3=(a+b)(a^2-ab+b^2)$ or by $a^3-b^3=(a-b)(a^2+ab+b^2)$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
(m+1)(m^2-m+1)[(m)^3-(1)^3]
\\\\=
(m+1)(m^2-m+1)[(m-1)(m^2+m+1)]
\\\\=
(m+1)(m^2-m+1)(m-1)(m^2+m+1)
\\\\=
(m+1)(m-1)(m^2-m+1)(m^2+m+1)
.\end{array}
