Answer
$${\left( {7k - 9j} \right)^4} = 2401{k^4} - 12348{k^3}j + 23814{k^2}{j^2} - 20412k{j^3} + 6561{j^4}$$
Work Step by Step
$$\eqalign{
& {\left( {7k - 9j} \right)^4} \cr
& {\left( {7k - 9j} \right)^4} = {\left( {7k + \left( { - 9j} \right)} \right)^4} \cr
& {\text{Apply the binomial theorem}} \cr
& {\left( {7k - 9j} \right)^4} = {\left( {7k} \right)^4} + \left( {4{\bf{C}}{\text{1}}} \right){\left( {7k} \right)^3}\left( { - 9j} \right) + \left( {4{\bf{C}}2} \right){\left( {7k} \right)^2}{\left( { - 9j} \right)^2} + \left( {4{\bf{C}}3} \right)\left( {7k} \right){\left( { - 9j} \right)^3}\, \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + {\left( { - 9j} \right)^4} \cr
& {\text{Evaluate each binomialcoefficient use }}\left( {c{\bf{C}}r} \right) = \frac{{n!}}{{\left( {n - r} \right)!r!}} \cr
& {\left( {7k - 9j} \right)^4} = {\left( {7k} \right)^4} + \frac{{4!}}{{3!1!}}{\left( {7k} \right)^3}\left( { - 9j} \right) + \frac{{4!}}{{2!2!}}{\left( {7k} \right)^2}{\left( { - 9j} \right)^2} + \frac{{4!}}{{1!3!}}\left( {7k} \right){\left( { - 9j} \right)^3}\, \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + {\left( { - 9j} \right)^4} \cr
& {\text{Simplify}} \cr
& {\left( {7k - 9j} \right)^4} = {\left( {7k} \right)^4} + 4{\left( {7k} \right)^3}\left( { - 9j} \right) + 6{\left( {7k} \right)^2}{\left( { - 9j} \right)^2} + 4\left( {7k} \right){\left( { - 9j} \right)^3}\, \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + {\left( { - 9j} \right)^4} \cr
& {\left( {7k - 9j} \right)^4} = 2401{k^4} - 12348{k^3}j + 23814{k^2}{j^2} - 20412k{j^3} + 6561{j^4} \cr} $$
