Solve the recurrence relation $a_n=3a_{n-1}+n^2-3$, with $a_0=1$. Solve the recurrence relation $a_n=3a_{n-1}+n^2-3$, with $a_0=1$.
My solutions: the homogeneous portion is $a_n=c3^n$, and the inhomogeneous portion is $a^*_n=-1/2n^2-3/4n+9/8$.
This results in a final recurrence relation of 
$$a_n=-\frac{1}{8}3^n-\frac{1}{2}n^2-\frac{3}{4}n+\frac{9}{8}.$$
I am just wondering if someone could check my work to make sure I have the procedure correct.

Well you've made a mistake,I'm not sure where is your mistake anyway here's is a different solution
$$a_n=3a_{n-1}+n^2-3\\a_{n-1}=3a_{n-2}+n^2-2n-2\\a_n-a_{n-1}=3a_{n-1}+n^2-3-3a_{n-2}-n^2+2n+2\\a_n=4a_{n-1}-3a_{n-2}+2n-1\\a_{n-1}=4a_{n-2}-3a_{n-3}+2n-3\\a_n-a_{n-1}=4a_{n-1}-3a_{n-2}+2n-1-4a_{n-2}+3a_{n-3}-2n+3\\a_n=5a_{n-1}-7a_{n-2}+3a_{n-3}+2\\a_n=5a_{n-1}-7a_{n-2}+3a_{n-3}+2\\a_{n-1}=5a_{n-2}-7a_{n-3}+3a_{n-4}+2\\a_n-a_{n-1}=5a_{n-1}-7a_{n-2}+3a_{n-3}+2-5a_{n-2}+7a_{n-3}-3a_{n-4}-2\\a_n=6a_{n-1}-12a_{n-2}+10a_{n-3}-3a_{n-4}\\t^4-6t+12t-10t+3=0\\t_{1,2,3}=1\\t_4=3\\a_n=C_1+nC_2+n^2C_3+C_43^n\\1=C_1+C_4\\a_1=1,a_2=4,a_3=18\\a_1=C_1+C_2+C_3+3C_4\\a_2=C_1+2C_2+4C_3+9C_4\\a_3=C_1+3C_2+9C_3+27C_4\\C_1=0,C_2=-\frac{3}{2},C_3=-\frac{1}{2},C_4=1\\a_n=-\frac{3}{2}n-\frac{1}{2}n^2+3^n$$
