Express \(\frac{1}{x^{3}-1}\) in partial fractions

\(\frac{1}{x^{3} - 1}\)

\(x^{3} - 1 = (x - 1)(x^{2} + x + 1)\)

\(\frac{1}{x^{3} - 1} = \frac{A}{x - 1} + \frac{Bx + C}{x^{2} + x + 1}\)

\(\frac{1}{x^{3} - 1} = \frac{A(x^{2} + x + 1) + (Bx + C)(x - 1)}{x^{3} - 1}\)

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): \(3A = 1 \implies A = \frac{1}{3}\)

\(A = -B \implies B = -\frac{1}{3}\)

\(C = A - 1 \implies C = \frac{1}{3} - 1 = -\frac{2}{3}\)

\(\therefore \frac{1}{x^{3} - 1} = \frac{1}{3(x - 1)} - \frac{x - 2}{3(x^{2} + x + 1)}\)

= \(\frac{1}{3}(\frac{1}{x - 1} - \frac{(x + 2)}{x^{2} + x + 1})\)

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