Finding particular solution of recurrence relation

Question

I am trying to solve recurrence relation problem -

$a_r - 3a_{r-1} + 3a_{r-2} + a_{r-3} = 4.$

So far I simplified it to $a_3 - 3a_2 + 3a - 3 = 0.$

Then, $a_2(a-3) + 3(a-1) = 0.$

But it seems I am going wrong somewhere. Is there any other way to solve this?

Answer 1

Assuming a constant solution $c$, we have

$$c-3c+3c+c=4.$$