Resolve \( \frac{1-x-x^{2}}{(1-x)\left(1-x^{3}\right)} \) into partial fractions.

To resolve the expression \( \frac{1-x-x^{2}}{(1-x)(1-x^{3})} \) into partial fractions, we start by factoring the denominator. The factor \( 1 - x^3 \) can be factored using the identity \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \). Thus, we can write: \[ 1 - x^3 = (1 - x)(1 + x + x^2). \] The complete factored form of the denominator is: \[ (1-x)(1-x^3) = (1-x)(1-x)(1+x+x^2) = (1-x)^2(1+x+x^2). \] Now we can set up the partial fraction decomposition: \[ \frac{1-x-x^{2}}{(1-x)^{2}(1+x+x^{2})} = \frac{A}{1-x} + \frac{B}{(1-x)^{2}} + \frac{Cx + D}{1+x+x^{2}}. \] Now, we multiply both sides by the denominator \( (1-x)^{2}(1+x+x^{2}) \) to eliminate the denominator: \[ 1 - x - x^2 = A(1-x)(1+x+x^2) + B(1+x+x^2) + (Cx+D)(1-x)^2. \] Next, we need to expand the right-hand side and collect like terms: 1. Expand \( A(1-x)(1+x+x^2) \): \[ A(1 - x + x + x^2 - x^2) = A(1 - x^2). \] 2. Expand \( B(1+x+x^2) \): \[ B + Bx + Bx^2. \] 3. Expand \( (Cx + D)(1-x)^2 \): \[ (Cx + D)(1 - 2x + x^2) = (Cx + D)(1 - 2x + x^2) = Cx - 2Cx^2 + Cx^3 + D - 2Dx + Dx^2. \] Combining all parts gives us: \[ 1 - x - x^2 = (A+D) + (B - 2D - C)x + (B + C - 2A)x^2 + Cx^3. \] Now we equate coefficients for \( x^0, x^1, x^2, x^3 \): 1. For \( x^0 \): \( 1 = A + D \) 2. For \( x^1 \): \( -1 = B - 2D - C \) 3. For \( x^2 \): \( -1 = B + C - 2A \) 4. For \( x^3 \): The coefficient is \( 0 = C \). Setting \( C = 0 \) in the equations, we can simplify: From \( C = 0 \): \\ The equation for \( x^0 \): \( 1 = A + D \) → \( D = 1 - A \). \\ The equation for \( x^1 \): \( -1 = B - 2(1 - A) \) → \( B + 2A = 1 \). \\ The equation for \( x^2 \): \( -1 = B - 2A \). Now substituting \( D = 1 - A \) into the second equation \( B + 2A = 1 \): Rearranging gives \( B = 1 - 2A \), and substituting this back into the last equation: \[ -1 = (1 - 2A) - 2A, \] which simplifies to \( -1 = 1 - 4A \) leading to: \[ 4A = 2 \implies A = \frac{1}{2}. \] So from \( D = 1 - A = \frac{1}{2} \) and substituting \( A = \frac{1}{2} \) into \( B + 2(1/2) = 1 \): \[ B + 1 = 1 \implies B = 0. \] In summary, we have: \[ A = \frac{1}{2}, B = 0, C = 0, D = \frac{1}{2}. \] Thus, the partial fraction decomposition is: \[ \frac{1-x-x^{2}}{(1-x)^{2}(1+x+x^{2