To find the term independent of x, we need the power of x to be zero in the expansion of \(\left( 4x^3 + \frac{1}{2x} \right)^8\).

Consider the general term in the binomial expansion:

\(T_r = \binom{8}{r} (4x^3)^{8-r} \left( \frac{1}{2x} \right)^r\).

Simplifying, we get:

\(T_r = \binom{8}{r} \cdot 4^{8-r} \cdot x^{3(8-r)} \cdot \frac{1}{2^r} \cdot x^{-r}\).

Combine the powers of x:

\(x^{3(8-r) - r} = x^{24 - 4r}\).

We want the power of x to be zero:

\(24 - 4r = 0\).

Solving for r gives:

\(r = 6\).

Substitute \(r = 6\) back into the expression for the term:

\(T_6 = \binom{8}{6} \cdot 4^2 \cdot \frac{1}{2^6}\).

Calculate \(\binom{8}{6} = 28\), \(4^2 = 16\), and \(2^6 = 64\).

Thus, \(T_6 = 28 \cdot 16 \cdot \frac{1}{64} = 28\).

Therefore, the term independent of x is 28.