Answer
$${\left( {3z - 5w} \right)^3} = 27{z^3} - 135{z^2} + 225{w^2}z - 125{w^3}$$
Work Step by Step
$$\eqalign{
& {\left( {3z - 5w} \right)^3} \cr
& {\left( {3z - 5w} \right)^3} = {\left( {3z + \left( { - 5w} \right)} \right)^3} \cr
& {\rm{Apply\, the\, binomial\, theorem}} \cr
& {\left( {3z - 5w} \right)^3} = {\left( {3z} \right)^3} + \left( \matrix{
3 \hfill \cr
1 \hfill \cr} \right){\left( {3z} \right)^2}\left( { - 5w} \right) + \left( \matrix{
3 \hfill \cr
2 \hfill \cr} \right)\left( {3z} \right){\left( { - 5w} \right)^2} + {\left( { - 5w} \right)^3}\, \cr
& {\rm{Evaluate\, each\, binomial\,coefficient \,use }}\left( \matrix{
n \hfill \cr
r \hfill \cr} \right) = {{n!} \over {\left( {n - r} \right)!r!}} \cr
& {\left( {3z - 5w} \right)^3} = {\left( {3z} \right)^3} + {{3!} \over {2!1!}}{\left( {3z} \right)^2}\left( { - 5w} \right) + {{3!} \over {1!2!}}\left( {3z} \right){\left( { - 5w} \right)^2} + {\left( { - 5w} \right)^3}\, \cr
& {\rm{Simplify}} \cr
& {\left( {3z - 5w} \right)^3} = 27{z^3} - 135{z^2} + 225{w^2}z - 125{w^3} \cr} $$
