Chứng minh rằng với mọi n nguyên dương ta có: a) n³+2n chia hết cho 3 b) 6 ^2n +3^n+2 + 3^n chia hết cho 11

$$\eqalign{ & a)\,\,A = {n^3} + 2n \cr & TH1:\,\,n = 3k\,\,\left( {k \in {N^*}} \right) \cr & A = {\left( {3k} \right)^3} + 2.3k\,\, \vdots \,\,3 \cr & TH2:\,\,n = 3k + 1 \cr & \Rightarrow A = {\left( {3k + 1} \right)^3} + 2.\left( {3k + 1} \right) \cr & \Rightarrow A = 27{k^3} + 27{k^2} + 9k + 1 + 6k + 2 \cr & \Rightarrow A = 27{k^3} + 27{k^2} + 15k + 3\,\,{\kern 1pt} \vdots \,\,3 \cr & TH3:\,\,n = 3k + 2 \cr & \Rightarrow A = {\left( {3k + 2} \right)^2} + 2\left( {3k + 2} \right) \cr & \Rightarrow A = 27{k^3} + 54{k^2} + 36k + 8 + 6k + 4 \cr & \Rightarrow A = 27{k^3} + 54{k^2} + 42k + 12\,\, \vdots \,\,3 \cr & Vay\,\,A\,\, \vdots \,\,3\,\,\forall n \in {N^*}. \cr} $$