12. If \( \frac{x+1}{(x-1)^{3}}=\frac{A}{x-1}+\frac{B}{(x-1)^{2}}+\frac{C}{(x-1)^{3}} \), then \( C= \) \( \begin{array}{llll}\text { a) } 2 & \text { b) } A+2 B & \text { c) } 2 B-A & \text { d) } 0\end{array} \)

To solve for \( C \) in the equation \[ \frac{x+1}{(x-1)^{3}} = \frac{A}{x-1} + \frac{B}{(x-1)^{2}} + \frac{C}{(x-1)^{3}}, \] we start by rewriting the right side with a common denominator: \[ \frac{A(x-1)^{2} + B(x-1) + C}{(x-1)^{3}}. \] Now we have: \[ \frac{x+1}{(x-1)^{3}} = \frac{A(x-1)^{2} + B(x-1) + C}{(x-1)^{3}}. \] Equating the numerators gives us: \[ x + 1 = A(x-1)^{2} + B(x-1) + C. \] Expanding the right-hand side, we get: \[ A(x^2 - 2x + 1) + B(x - 1) + C = Ax^2 + (B - 2A)x + (A + C - B). \] To find \( C \), we compare coefficients with the left-hand side, \( x + 1 \): 1. The coefficient of \( x^2 \): \( A = 0 \). 2. The coefficient of \( x \): \( B - 2A = 1 \) which simplifies to \( B = 1 \) since \( A = 0 \). 3. The constant term: \( A + C - B = 1 \), substituting \( A = 0 \) and \( B = 1 \), we have \( C - 1 = 1 \) so \( C = 2 \). Thus, the value of \( C \) is \( 2 \). The correct answer is \( \text{a) } 2 \).