2 Decompose \[ \frac{x^{3}-x+2}{x^{3}-1} \] into partial fractions.

To decompose \(\frac{x^{3}-x+2}{x^{3}-1}\) into partial fractions, we first factor the denominator \(x^{3} - 1\). We can use the formula for the difference of cubes: \[ x^{3} - 1 = (x - 1)(x^2 + x + 1) \] So, we rewrite the expression: \[ \frac{x^{3} - x + 2}{(x - 1)(x^{2} + x + 1)} \] Next, we set up the partial fraction decomposition: \[ \frac{x^{3} - x + 2}{(x - 1)(x^{2} + x + 1)} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + x + 1} \] Where \(A\), \(B\), and \(C\) are constants we need to determine. Multiplying through by the denominator \((x - 1)(x^{2} + x + 1)\) gives: \[ x^{3} - x + 2 = A(x^2 + x + 1) + (Bx + C)(x - 1) \] Expanding the right-hand side: \[ A(x^2 + x + 1) + (Bx^2 - Bx + Cx - C) = (A + B)x^2 + (A - B + C)x + (A - C) \] Now, we equate coefficients for the terms from both sides: 1. For \(x^3\): Coefficient is 1 (from \(x^{3}\)) on the left-hand side, but 0 from our setup because there’s no \(x^3\) term on the right side, so we do not have any equation from this. 2. For \(x^2\): \(A + B = 0\) 3. For \(x^1\): \(A - B + C = -1\) 4. For the constant term: \(A - C = 2\) Now we have a system of equations: 1. \(A + B = 0\) (1) 2. \(A - B + C = -1\) (2) 3. \(A - C = 2\) (3) From (1), we get \(B = -A\). Substituting \(B = -A\) into (2): \[ A - (-A) + C = -1 \quad \Rightarrow \quad 2A + C = -1 \quad \Rightarrow \quad C = -1 - 2A \quad \text{(4)} \] Now substituting \(C\) from (4) into (3): \[ A - (-1 - 2A) = 2 \quad \Rightarrow \quad A + 1 + 2A = 2 \quad \Rightarrow \quad 3A = 1 \quad \Rightarrow \quad A = \frac{1}{3} \] Then substituting back to find \(B\) and \(C\): \[ B = -A = -\frac{1}{3} \] Substituting \(A\) into (4): \[ C = -1 - 2\left(\frac{1}{3}\right) = -1 - \frac{2}{3} = -\frac{5}{3} \] Now we have: \[ A = \frac{1}{3}, \quad B = -\frac{1}{3}, \quad C = -\frac{5}{3} \] Therefore, we can write the partial fraction decomposition as: \[ \frac{x^{3}-x+2}{(x - 1)(x^{2} + x + 1)} = \frac{1/3}{x - 1} + \frac{-\frac{1}{3}x - \frac{5}{3}}{x^{2} + x + 1} \] Or \[ = \frac{1}{3(x - 1)} - \frac{1}{3}\frac{x + 5}{x^{2} + x + 1} \] That's the final decomposition!