To express \(\frac{1}{x^{3}-1}\) in partial fractions, we first need to factor the denominator. The expression \(x^3 - 1\) can be factored using the difference of cubes formula: \[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \] In our case, \(a = x\) and \(b = 1\), so we have: \[ x^3 - 1 = (x - 1)(x^2 + x + 1) \] Now, we can express \(\frac{1}{x^3 - 1}\) as: \[ \frac{1}{x^3 - 1} = \frac{1}{(x - 1)(x^2 + x + 1)} \] Next, we set up the partial fraction decomposition: \[ \frac{1}{(x - 1)(x^2 + x + 1)} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + x + 1} \] Here, \(A\), \(B\), and \(C\) are constants that we need to determine. To combine the right-hand side into a single fraction, we find a common denominator: \[ \frac{A(x^2 + x + 1) + (Bx + C)(x - 1)}{(x - 1)(x^2 + x + 1)} \] Now, we equate the numerators: \[ 1 = A(x^2 + x + 1) + (Bx + C)(x - 1) \] Expanding the right-hand side: \[ 1 = A(x^2 + x + 1) + Bx^2 - Bx + Cx - C \] Combining like terms gives us: \[ 1 = (A + B)x^2 + (A - B + C)x + (A - C) \] Now, we can set up a system of equations by comparing coefficients from both sides: 1. For \(x^2\): \(A + B = 0\) 2. For \(x\): \(A - B + C = 0\) 3. For the constant term: \(A - C = 1\) Now, we can solve this system step by step. From the first equation, we can express \(B\) in terms of \(A\): \[ B = -A \] Substituting \(B = -A\) into the second equation: \[ A - (-A) + C = 0 \implies 2A + C = 0 \implies C = -2A \] Now substituting \(C = -2A\) into the third equation: \[ A - (-2A) = 1 \implies A + 2A = 1 \implies 3A = 1 \implies A = \frac{1}{3} \] Now substituting \(A = \frac{1}{3}\) back to find \(B\) and \(C\): \[ B = -A = -\frac{1}{3} \] \[ C = -2A = -2 \cdot \frac{1}{3} = -\frac{2}{3} \] Now we have: - \(A = \frac{1}{3}\) - \(B = -\frac{1}{3}\) - \(C = -\frac{2}{3}\) Thus, the partial fraction decomposition is: \[ \frac{1}{(x - 1)(x^2 + x + 1)} = \frac{\frac{1}{3}}{x - 1} + \frac{-\frac{1}{3}x - \frac{2}{3}}{x^2 + x + 1} \] This can be rewritten as: \[ \frac{1}{3}\left(\frac{1}{x - 1} - \frac{x + 2}{x^2 + x + 1}\right) \] Now, looking at the options provided: - **Option A**: \(\frac{1}{3}(\frac{1}{x - 1} - \frac{(x + 2)}{x^{2} + x + 1})\) is correct. - **Option B**: \(\frac{1}{3}(\frac{1}{x - 1} - \frac{x - 2}{x^{2} + x + 1})\) is incorrect because it has the wrong numerator for the second term. - **Option C**: \(\frac{1}{3}(\frac{1}{x - 1} - \frac{(x - 2)}{x^{2} + x + 1})\) is also incorrect for the same reason as option B. - **Option D**: \(\frac{1}{3}(\frac{1}{x - 1} - \frac{(x - 1)}{x^{2} - x - 1})\) is incorrect because it has a different denominator and incorrect numerator. ### Summary: - The correct option is **A**: \(\frac{1}{3}(\frac{1}{x - 1} - \frac{(x + 2)}{x^{2} + x + 1})\). - The denominator \(x^3 - 1\) factors into \((x - 1)(