Answer
$${m^{12}} - 4{m^7} + 6{m^2} - 4{m^{ - 3}} + {m^{ - 8}}$$
Work Step by Step
$$\eqalign{
& {\left( {{m^3} - {m^{ - 2}}} \right)^4} = {\left( {{m^3} + \left( { - {m^{ - 2}}} \right)} \right)^4} \cr
& {\rm{Apply\, the\, binomial\, theorem}} \cr
& {\left( {{m^3} - {m^{ - 2}}} \right)^4} = {\left( {{m^3}} \right)^4} + \left( \matrix{
4 \hfill \cr
1 \hfill \cr} \right){\left( {{m^3}} \right)^3}\left( { - {m^{ - 2}}} \right) + \left( \matrix{
4 \hfill \cr
2 \hfill \cr} \right){\left( {{m^3}} \right)^2}{\left( { - {m^{ - 2}}} \right)^2} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \left( \matrix{
4 \hfill \cr
3 \hfill \cr} \right)\left( {{m^3}} \right){\left( { - {m^{ - 2}}} \right)^3}\,\, + {\left( { - {m^{ - 2}}} \right)^4} \cr
& {\rm{Evaluate\, each\, binomial\,coefficient\, use }}\left( \matrix{
c \hfill \cr
r \hfill \cr} \right) = {{n!} \over {\left( {n - r} \right)!r!}} \cr
& {\left( {{m^3} - {m^{ - 2}}} \right)^4} = {\left( {{m^3}} \right)^4} + {{4!} \over {3!1!}}{\left( {{m^3}} \right)^3}\left( { - {m^{ - 2}}} \right) + {{4!} \over {2!2!}}{\left( {{m^3}} \right)^2}{\left( { - {m^{ - 2}}} \right)^2} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + {{4!} \over {1!3!}}\left( {{m^3}} \right){\left( { - {m^{ - 2}}} \right)^3}\, + {\left( { - {m^{ - 2}}} \right)^4} \cr
& {\rm{Simplify}} \cr
& {\left( {{m^3} - {m^{ - 2}}} \right)^4} = {\left( {{m^3}} \right)^4} + 4{\left( {{m^3}} \right)^3}\left( { - {m^{ - 2}}} \right) + 6{\left( {{m^3}} \right)^2}{\left( { - {m^{ - 2}}} \right)^2} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 4\left( {{m^3}} \right){\left( { - {m^{ - 2}}} \right)^3}\,\, + {\left( { - {m^{ - 2}}} \right)^4} \cr
& {\left( {{m^3} - {m^{ - 2}}} \right)^4} = {m^{12}} - 4{m^7} + 6{m^2} - 4{m^{ - 3}} + {m^{ - 8}} \cr} $$
