To find the term independent of x, we use the binomial expansion formula:
\(\left( a + b \right)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\).
Here, \(a = 2x\) and \(b = \frac{1}{2x^3}\), and \(n = 8\).
We need the term where the power of x is zero. The general term is:
\(T_k = \binom{8}{k} (2x)^{8-k} \left( \frac{1}{2x^3} \right)^k\).
Simplifying, we get:
\(T_k = \binom{8}{k} \cdot 2^{8-k} \cdot x^{8-k} \cdot \frac{1}{2^k} \cdot x^{-3k}\).
\(T_k = \binom{8}{k} \cdot 2^{8-2k} \cdot x^{8-4k}\).
We want the power of x to be zero:
\(8 - 4k = 0\).
Solving for \(k\), we get \(k = 2\).
Substitute \(k = 2\) into the term:
\(T_2 = \binom{8}{2} \cdot 2^{8-4} \cdot x^{0}\).
\(T_2 = 28 \cdot 2^4 \cdot 1\).
\(T_2 = 28 \cdot 16\).
\(T_2 = 448\).
Thus, the term independent of x is 448.
