Step-by-step Solution:

Let's find the coefficient of \( x^{10} \) in both terms separately. First expansion Term: \( \left(x^2 + \frac{1}{x}\right)^{12} \) The general term is given by \( T_{k+1} = \binom{12}{k} (x^2)^{12-k} (x^{-1})^k = \binom{12}{k} x^{24-2k-k} = \binom{12}{k} x^{24-3k} \). For the coefficient of \( x^{10} \), we set the exponent to 10: \[ 24 - 3k = 10 \implies 3k = 14 \] Since \( k \) must be an integer, there is no valid \( x^{10} \) term in this first binomial expansion. Second expansion Term: \( \left(x + \frac{1}{x^2}\right)^{12} \) The general term is given by \( T_{m+1} = \binom{12}{m} (x)^{12-m} (x^{-2})^m = \binom{12}{m} x^{12-m-2m} = \binom{12}{m} x^{12-3m} \). For the coefficient of \( x^{10} \), we set the exponent to 10: \[ 12 - 3m = 10 \implies 3m = 2 \] Since \( m \) must also be an integer, there is no \( x^{10} \) term in the second expansion either. Since neither expansion yields an \( x^{10} \) term, the overall coefficient of \( x^{10} \) in their sum evaluates to 0.