$$\eqalign{
& a)\,\,{7^{2018}} + {7^{2019}} + {7^{2020}} \cr
& = {7^{2018}}\left( {1 + 7 + {7^2}} \right) = {7^{2018}}.57\,\, \vdots \,\,57 \cr
& b)\,\,{5^{100}} + {5^{98}} \cr
& = {5^{98}}\left( {{5^2} + 1} \right) = {5^{98}}.26 \cr
& = {5^{98}}.2.13\,\, \vdots \,\,13 \cr
& c)\,\,Chua\,\,ro\,\,de\,\,bai \cr
& d)\,{\left( {2n - 1} \right)^3} - \left( {2n - 1} \right) \cr
& = \left( {2n - 1} \right)\left[ {{{\left( {2n - 1} \right)}^2} - 1} \right] \cr
& = \left( {2n - 1} \right)\left( {2n - 1 - 1} \right)\left( {2n - 1 + 1} \right) \cr
& = 2n\left( {2n - 1} \right)\left( {2n - 2} \right) \cr
& = 4\left( {n - 1} \right)n\left( {2n - 1} \right) \cr
& \left( {n - 1} \right)n\,\, \vdots \,\,2\,\,\forall n \Rightarrow 4\left( {n - 1} \right)n\,\,\, \vdots \,\,8\,\,\forall n \cr
& \Rightarrow 4\left( {n - 1} \right)n\left( {2n - 1} \right)\,\, \vdots \,\,8\,\,\forall n \cr} $$
