Answer
$${\left( {{1 \over k} + \sqrt 3 p} \right)^3} = {1 \over {{k^3}}} + {{3\sqrt 3 p} \over {{k^2}}} + {{9{p^2}} \over k} + 3\sqrt 3 {p^3}\,$$
Work Step by Step
$$\eqalign{
& {\left( {{1 \over k} + \sqrt 3 p} \right)^3} \cr
& {\rm{Apply\, the\, binomial\, theorem}} \cr
& {\left( {{1 \over k} + \sqrt 3 p} \right)^3} = {\left( {{1 \over k}} \right)^3} + \left( \matrix{
3 \hfill \cr
1 \hfill \cr} \right){\left( {{1 \over k}} \right)^2}\left( {\sqrt 3 p} \right) + \left( \matrix{
3 \hfill \cr
2 \hfill \cr} \right)\left( {{1 \over k}} \right){\left( {\sqrt 3 p} \right)^2} + {\left( {\sqrt 3 p} \right)^3}\, \cr
& {\rm{Evaluate\,each\, binomial\,coefficient\, use }}\left( \matrix{
n \hfill \cr
r \hfill \cr} \right) = {{n!} \over {\left( {n - r} \right)!r!}} \cr
& {\left( {{1 \over k} + \sqrt 3 p} \right)^3} = {\left( {{1 \over k}} \right)^3} + {{3!} \over {2!1!}}{\left( {{1 \over k}} \right)^2}\left( {\sqrt 3 p} \right) + {{3!} \over {1!2!}}\left( {{1 \over k}} \right){\left( {\sqrt 3 p} \right)^2} + {\left( {\sqrt 3 p} \right)^3}\, \cr
& {\rm{Simplify}} \cr
& {\left( {{1 \over k} + \sqrt 3 p} \right)^3} = {\left( {{1 \over k}} \right)^3} + 3{\left( {{1 \over k}} \right)^2}\left( {\sqrt 3 p} \right) + 3\left( {{1 \over k}} \right){\left( {\sqrt 3 p} \right)^2} + {\left( {\sqrt 3 p} \right)^3}\, \cr
& {\left( {{1 \over k} + \sqrt 3 p} \right)^3} = {1 \over {{k^3}}} + {{3\sqrt 3 p} \over {{k^2}}} + {{9{p^2}} \over k} + 3\sqrt 3 {p^3}\, \cr} $$
