<s>[INST] Find the recursive formula for $a_{n}={b^n}$ for $n>2$ How can I solve it,
and can anybody help me to find recursive formula for $a_{n}=3n^3$ for $n\geq 0$.
 [/INST] Set $b_n=a_{n+1}-a_n=3(n+1)^3-3n^3=9n^2+9n+3$ 
Set $c_n=b_{n+1}-b_n=9(n+1)^2+9(n+1)+3-(9n^2+9n+3)=18n+18$
So $c_{n+1}=c_n+18$ and $c_0=18$
Then $b_{n+2}-b_{n+1}=b_{n+1}-b_n+18$ 
then $a_{n+3}-a_{n+2}-(a_{n+2}-a_{n+1})=a_{n+2}-a_{n+1}-(a_{n+1}-a_n)+18$
So $a_{n+3}=3a_{n+2}-3a_{n+1}+a_n+18$, with $a_0=0, a_1=3, a_2=24$
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