#### Answer

$(m-n)(m^2+mn+n^2+m+n)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
Group the terms of the given expression, $
m^3+m^2-n^3-n^2
,$ such that the factored form of the groupings will result to a factor that is common to the entire expression. Then, use the factoring of $2$ cubes and the factoring of the difference of $2$ squares.
$\bf{\text{Solution Details:}}$
Grouping the first and third terms and the second and fourth terms, the given expression is equivalent to
\begin{array}{l}\require{cancel}
(m^3-n^3)+(m^2-n^2)
.\end{array}
The expressions $
m^3
$ and $
n^3
$ are both perfect cubes (the cube root is exact). Hence, $
m^3-n^3
$ 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)^3-(n)^3]+(m^2-n^2)
\\\\=
(m-n)[(m)^2+m(n)+(n)^2]+(m^2-n^2)
\\\\=
(m-n)(m^2+mn+n^2)+(m^2-n^2)
.\end{array}
The expressions $
m^2
$ and $
n^2
$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $
m^2-n^2
,$ 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-n)(m^2+mn+n^2)+[(m)^2-(n)^2]
\\\\=
(m-n)(m^2+mn+n^2)+(m-n)(m+n)
.\end{array}
Factoring the $GCF=
(m-n)
$ of the entire expression above results to
\begin{array}{l}\require{cancel}
(m-n)[(m^2+mn+n^2)+(m+n)]
\\\\
(m-n)(m^2+mn+n^2+m+n)
.\end{array}