Determine the value of each of the constants: \( A, B \) and \( C \). \[ \left(\mathbf{1}+x-x^{2}\right)^{6}=\mathbf{1}+A x+B x^{2}+C x^{3}+\cdots \]

To find the constants \( A, B, \) and \( C \) in the expansion of \( \left( \mathbf{1} + x - x^2 \right)^6 \), we can use the multinomial expansion. The expression can be rewritten as: \[ \left( \mathbf{1} + x - x^2 \right)^6 = \sum_{k_1 + k_2 + k_3 = 6} \frac{6!}{k_1! k_2! k_3!} \cdot 1^{k_1} \cdot x^{k_2} \cdot (-x^2)^{k_3} \] This simplifies to: \[ \sum_{k_1 + k_2 + k_3 = 6} \frac{6!}{k_1! k_2! k_3!} \cdot x^{k_2 + 2k_3} \cdot (-1)^{k_3} \] We need to find the coefficients of \( x^1 \), \( x^2 \), and \( x^3 \). ### Finding \( A \) (coefficient of \( x^1 \)) For \( x^1 \), we need \( k_2 + 2k_3 = 1 \). The only possible values are: - \( k_2 = 1 \) and \( k_3 = 0 \) (thus \( k_1 = 5 \)) Calculating the coefficient: \[ \frac{6!}{5!1!0!} \cdot (-1)^0 = \frac{720}{120 \cdot 1 \cdot 1} = 6 \] Thus, \( A = 6 \). ### Finding \( B \) (coefficient of \( x^2 \)) For \( x^2 \), we need \( k_2 + 2k_3 = 2 \). The possible combinations are: 1. \( k_2 = 2, k_3 = 0 \) (thus \( k_1 = 4 \)) 2. \( k_2 = 0, k_3 = 1 \) (thus \( k_1 = 5 \)) Calculating the coefficients for each case: 1. For \( k_2 = 2, k_3 = 0 \): \[ \frac{6!}{4!2!0!} \cdot (-1)^0 = \frac{720}{24 \cdot 2 \cdot 1} = 15 \] 2. For \( k_2 = 0, k_3 = 1 \): \[ \frac{6!}{5!0!1!} \cdot (-1)^1 = \frac{720}{120 \cdot 1 \cdot 1} \cdot (-1) = -6 \] Adding these contributions together: \[ B = 15 - 6 = 9 \] ### Finding \( C \) (coefficient of \( x^3 \)) For \( x^3 \), we need \( k_2 + 2k_3 = 3 \). The possible combinations are: 1. \( k_2 = 3, k_3 = 0 \) (thus \( k_1 = 3 \)) 2. \( k_2 = 1, k_3 = 1 \) (thus \( k_1 = 4 \)) 3. \( k_2 = 0, k_3 = 2 \) (thus \( k_1 = 4 \)) Calculating the coefficients for each case: 1. For \( k_2 = 3, k_3 = 0 \): \[ \frac{6!}{3!3!0!} \cdot (-1)^0 = \frac{720}{6 \cdot 6 \cdot 1} = 20 \] 2. For \( k_2 = 1, k_3 = 1 \): \[ \frac{6!}{4!1!1!} \cdot (-1)^1 = \frac{720}{24 \cdot 1 \cdot 1} \cdot (-1) = -30 \] 3. For \( k_2 = 0, k_3 = 2 \): \[ \frac{6!}{4!0!2!} \cdot (-1)^2 = \frac{720}{24 \cdot 1 \cdot 2} = 15 \] Adding these contributions together: \[ C =