Complex Invariant

For any finite collection of complex numbers $a_1,a_2,...,a_n$ consider the quantity $X(a_1,a_2,...,a_n) \: = \; 3^n \prod_{j=1}^n (a_j - 1)$ Since $3ab - 3a - 3b + 4 = 3(a-1)(b-1) + 1$ , we deduce that $X(a_1,a_2,...,a_{n-2},a_{n-1},a_n) \; = \; X\big(a_1,a_2,...,a_{n-2},3a_{n-1}a_n - 3a_{n-1} - 3a_n + 4\big)$ Thus, if $a_1,a_2,...,a_n$ are a set of numbers, and $b_1,b_2,...,b_{n-1}$ are the set of numbers obtained when two of the original set are erased and replaced according to the rule given in the problem, then $X(a_1,a_2,...,a_n) = X(b_1,b_2,...,b_{n-1})$ .

Thus, if $c$ is the single number left after the fifth step of the given process, then $3(c-1) \; = \; X(c) \; = \; X(\zeta_0,\zeta_1,...,\zeta_5)$ where $\zeta_j \; (0 \le j \le 5)$ are the six roots of the equation $X^6 + 2 = 0$ . But then $\begin{aligned} X(\zeta_0,\zeta_1,...,\zeta_5) & = \; 3^6\prod_{j=0}^5 (\zeta_j - 1) \; = \; 3^6 \prod_{j=0}^5 (1 - \zeta_j) \; = \; 3^6 \big(X^6 + 2\big)\Big|_{X=1} \; = \; 3^6 \times 3 = 3^7 \end{aligned}$ and hence the final number $c = 3^6 + 1 = \boxed{730}$ , irrespective of the order in which the numbers are chosen to be deleted/replaced.