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

To find the constants \( A, B, \) and \( C \) in the expansion of \( (1+x-x^2)^6 \), we can use the multinomial expansion. The expression can be rewritten as: \[ (1 + (x - x^2))^6 \] We will expand this using the binomial theorem, which states that: \[ (1 + u)^n = \sum_{k=0}^{n} \binom{n}{k} u^k \] In our case, \( u = x - x^2 \) and \( n = 6 \). Thus, we have: \[ (1 + (x - x^2))^6 = \sum_{k=0}^{6} \binom{6}{k} (x - x^2)^k \] Next, we need to expand \( (x - x^2)^k \). Using the binomial theorem again, we have: \[ (x - x^2)^k = \sum_{j=0}^{k} \binom{k}{j} x^j (-x^2)^{k-j} = \sum_{j=0}^{k} \binom{k}{j} (-1)^{k-j} x^{j + 2(k-j)} = \sum_{j=0}^{k} \binom{k}{j} (-1)^{k-j} x^{2k - j} \] Now, we need to find the coefficients of \( x^1, x^2, \) and \( x^3 \) in the overall expansion. ### Coefficient of \( x^1 \) (i.e., \( A \)) To get \( x^1 \), we need \( 2k - j = 1 \), which gives \( j = 2k - 1 \). This means \( j \) must be odd, and \( k \) must be at least 1 (since \( j \) cannot be negative). The possible values of \( k \) are \( 1, 2, 3, 4, 5, 6 \). - For \( k = 1 \): \( j = 1 \) gives \( \binom{6}{1} \binom{1}{1} (-1)^0 = 6 \). - For \( k = 2 \): \( j = 3 \) gives \( \binom{6}{2} \binom{2}{1} (-1)^1 = 15 \cdot 2 \cdot (-1) = -30 \). - For \( k = 3 \): \( j = 5 \) gives \( \binom{6}{3} \binom{3}{1} (-1)^0 = 20 \cdot 3 = 60 \). - For \( k = 4 \): \( j = 7 \) is not possible. - For \( k = 5 \): \( j = 9 \) is not possible. - For \( k = 6 \): \( j = 11 \) is not possible. Thus, the coefficient of \( x^1 \) is: \[ A = 6 - 30 + 60 = 36 \] ### Coefficient of \( x^2 \) (i.e., \( B \)) To get \( x^2 \), we need \( 2k - j = 2 \), which gives \( j = 2k - 2 \). The possible values of \( k \) are \( 1, 2, 3, 4, 5, 6 \). - For \( k = 1 \): \( j = 0 \) gives \( \binom{6}{1} \binom{1}{0} (-1)^1 = -6 \). - For \( k = 2 \): \( j = 2 \) gives \( \binom{6}{2} \binom{2}{2} (-1)^0 = 15 \). - For \( k = 3 \): \( j = 4 \) gives \( \binom{6}{3} \binom{3}{1} (-1)^1 = 20 \cdot 3 \cdot (-1) = -60 \). - For \( k = 4 \): \( j = 6 \) gives \( \binom{6}{4} \binom{4}{2} (-1)^0 = 15 \cdot 6 = 90 \). - For \( k = 5 \): \( j = 8 \) is not possible. - For \( k = 6 \