Solve the recurrence an=6an−1−9an−2a_n = 6a_{n-1} - 9a_{n-2}an=6an−1−9an−2 with a0=2a_0 = 2a0=2 and a1=9a_1 = 9a1=9. What is a3a_3a3?
108
120
135
162