Answer
$${\rm{sixth\, term}} = - 3584{h^3}{j^5}$$
Work Step by Step
$$\eqalign{
& {\rm{sixth \,term\, of }}{\left( {4h - j} \right)^8} \cr
& {\rm{In\, the\, sixth\, term, }} - j{\rm{ \,has\, an\, exponent \,of }}6 - 1,{\rm{ or 5, while\, }}4h \cr
& {\rm{has\, an\, exponent\, of\, 8}} - 5,{\rm{ or 3}}{\rm{.}} \cr
& \cr
& {\rm{Using\, the\, }}k{\rm{th\, Term\, of\, The\, Binomial\, Expansion}} \cr
& {\rm{sixth term}} = \left( \matrix{
8 \hfill \cr
5 \hfill \cr} \right){\left( {4h} \right)^3}{\left( { - j} \right)^5} \cr
& {\rm{sixth\, term}} = {{8!} \over {3!5!}}\left( {64{h^3}} \right)\left( { - {j^5}} \right) \cr
& {\rm{sixth \,term}} = 56\left( {64{h^3}} \right)\left( { - {j^5}} \right) \cr
& {\rm{sixth\, term}} = - 3584{h^3}{j^5} \cr} $$
