To express the function \(\frac{1}{x^3 - 1}\) 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 find these constants, we multiply both sides by the denominator \((x - 1)(x^2 + x + 1)\): \[ 1 = A(x^2 + x + 1) + (Bx + C)(x - 1) \] Expanding the right-hand side: \[ 1 = A(x^2 + x + 1) + Bx^2 - Bx + Cx - C \] Combining like terms gives: \[ 1 = (A + B)x^2 + (A - B + C)x + (A - C) \] Now, we can equate the coefficients from both sides of the equation. Since the left side is \(1\) (which can be thought of as \(0x^2 + 0x + 1\)), we have the following system of equations: 1. \(A + B = 0\) (coefficient of \(x^2\)) 2. \(A - B + C = 0\) (coefficient of \(x\)) 3. \(A - C = 1\) (constant term) Now, we can solve this system step by step. From equation (1), we can express \(B\) in terms of \(A\): \[ B = -A \] Substituting \(B = -A\) into equation (2): \[ A - (-A) + C = 0 \implies 2A + C = 0 \implies C = -2A \] Now substituting \(C = -2A\) into equation (3): \[ A - (-2A) = 1 \implies A + 2A = 1 \implies 3A = 1 \implies A = \frac{1}{3} \] Now substituting \(A = \frac{1}{3}\) back to find \(B\) and \(C\): \[ B = -A = -\frac{1}{3} \] \[ C = -2A = -2 \cdot \frac{1}{3} = -\frac{2}{3} \] Thus, we have: \[ A = \frac{1}{3}, \quad B = -\frac{1}{3}, \quad C = -\frac{2}{3} \] Now we can write the partial fraction decomposition: \[ \frac{1}{(x - 1)(x^2 + x + 1)} = \frac{\frac{1}{3}}{x - 1} + \frac{-\frac{1}{3}x - \frac{2}{3}}{x^2 + x + 1} \] This can be simplified to: \[ \frac{1}{3} \left( \frac{1}{x - 1} - \frac{x + 2}{x^2 + x + 1} \right) \] Now, let's analyze the options provided: ### Option A: \[ \frac{1}{3} \left( \frac{1}{x - 1} - \frac{x + 2}{x^2 + x + 1} \right) \] This matches our derived expression. ### Option B: \[ \frac{1}{3} \left( \frac{1}{x - 1} - \frac{x - 2}{x^2 + x + 1} \right) \] This is incorrect because the numerator of the second term should be \(- (x + 2)\), not \(- (x - 2)\). ### Option C: \[ \frac{1}{3} \left( \frac{1}{x - 1} - \frac{x - 2}{x^2 + x + 1} \right) \] This is also incorrect for the same reason as Option B. ### Option D: \[ \frac{1}{3} \left( \frac{1}{x - 1} - \frac{x - 1}{x^2 - x - 1} \right) \] This is incorrect because the second term has the wrong denominator and numerator. ### Conclusion: The correct option is **A**. ### Revision Summary: - Factor the denominator \(x^3 - 1\) using the difference of cubes. - Set