\(\frac{1}{x^{3} - 1}\) \(x^{3} - 1 = (x - 1)(x^{2} + x + 1)\) \(\frac{1}{x^{3} - 1} = \frac{A}{x - 1} + \frac{Bx + C}{x^{2} + x + 1}\) \(\frac{1}{x^{3} - 1} = \frac{A(x^{2} + x + 1) + (Bx + C)(x - 1)}{x^{3} - 1}\) Comparing the two sides of the equation, \(A + B = 0 ... (1)\) \(A - B + C = 0 ... (2)\) \(A - C = 1 ... (3)\) From (3), \(C = A - 1\), putting that in (2), \(A - B = -C \implies A - B = 1 - A\) \(2A - B = 1 ... (4)\) (1) + (4): \(3A = 1 \implies A = \frac{1}{3}\) \(A = -B \implies B = -\frac{1}{3}\) \(C = A - 1 \implies C = \frac{1}{3} - 1 = -\frac{2}{3}\) \(\therefore \frac{1}{x^{3} - 1} = \frac{1}{3(x - 1)} - \frac{x - 2}{3(x^{2} + x + 1)}\) = \(\frac{1}{3}(\frac{1}{x - 1} - \frac{(x + 2)}{x^{2} + x + 1})\)