The characteristic equation of the recurrence relation is $r^3-r^2-6r-4=0$

Its roots are r=-1. Hence the sequence {$a_n$} is a solution to the recurrence relation if and only if

$$a_n=α_1 (-1)^n+-α_2 (n)(-1)^n+α_3 (n x n)(-1)^n=α_1+α_1n+α_3(n \times n)$$

For some constant $α_1$ and $α_2$.

From the initial condition, it follows that

Let $a_0=1, a_1=2, a_2=3 \\ a_0=1=α_1 + α_2(0) + α_3(0) \\ a_1= 2 = - α_1 - α_2 - α_3 \\ a_2= 3 = α_1 +2α_2 + 4 α_3$

Solving the equations, we get $α_1=1, α_2=-7, α_3=4$

Hence the solution is the sequence {$a_n$} with

$a_r =1 - 7n + 4n^2$