By considering the general term, determine the term independent of \( x \) in the expansion of \( \left(2 x^{3}-\frac{1}{4 x^{5}}\right)^{8} \)

To find the term independent of \( x \) in the expansion of \( \left(2 x^{3}-\frac{1}{4 x^{5}}\right)^{8} \), we can use the binomial theorem. The general term in the expansion of \( (a + b)^n \) is given by: \[ T_k = \binom{n}{k} a^{n-k} b^k \] In our case, we have \( a = 2x^3 \), \( b = -\frac{1}{4x^5} \), and \( n = 8 \). Thus, the general term \( T_k \) can be expressed as: \[ T_k = \binom{8}{k} (2x^3)^{8-k} \left(-\frac{1}{4x^5}\right)^k \] Now, we simplify \( T_k \): \[ T_k = \binom{8}{k} (2^{8-k} x^{3(8-k)}) \left(-\frac{1}{4^k x^{5k}}\right) \] This can be rewritten as: \[ T_k = \binom{8}{k} (-1)^k \frac{2^{8-k}}{4^k} x^{3(8-k) - 5k} \] Next, we simplify \( \frac{2^{8-k}}{4^k} \): \[ 4^k = (2^2)^k = 2^{2k} \quad \Rightarrow \quad \frac{2^{8-k}}{4^k} = \frac{2^{8-k}}{2^{2k}} = 2^{8-k-2k} = 2^{8-3k} \] Thus, we have: \[ T_k = \binom{8}{k} (-1)^k 2^{8-3k} x^{3(8-k) - 5k} \] Now, we need to find the exponent of \( x \): \[ 3(8-k) - 5k = 24 - 3k - 5k = 24 - 8k \] We want the term independent of \( x \), which means we set the exponent of \( x \) to zero: \[ 24 - 8k = 0 \] Solving for \( k \): \[ 8k = 24 \quad \Rightarrow \quad k = 3 \] Now, we substitute \( k = 3 \) back into the expression for \( T_k \): \[ T_3 = \binom{8}{3} (-1)^3 2^{8-3 \cdot 3} x^{24 - 8 \cdot 3} \] Calculating \( T_3 \): \[ T_3 = \binom{8}{3} (-1)^3 2^{8-9} x^{24 - 24} \] \[ = \binom{8}{3} (-1)^3 2^{-1} x^0 \] \[ = \binom{8}{3} (-1) \frac{1}{2} \] Now, we compute \( \binom{8}{3} \): \[ \binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56 \] Thus, we have: \[ T_3 = 56 \cdot (-1) \cdot \frac{1}{2} = -28 \] Therefore, the term independent of \( x \) in the expansion is: \[ \boxed{-28} \]