The recurrence Tn=2Tn−1+8T_n = 2T_{n-1} + 8Tn=2Tn−1+8 with T0=0T_0 = 0T0=0 has solution Tn=8(2n−1)T_n = 8(2^n - 1)Tn=8(2n−1). What is T3T_3T3?
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