To find the term independent of x, we use the binomial expansion formula:
\(\left( x - \frac{3}{2x} \right)^6 = \sum_{k=0}^{6} \binom{6}{k} x^{6-k} \left( -\frac{3}{2x} \right)^k\)
The term independent of x occurs when the powers of x cancel out, i.e., when:
\(6-k-k = 0\)
Solving for k, we get:
\(6 - 2k = 0\)
\(k = 3\)
\(Substitute k = 3 into the binomial term:\)
\(\binom{6}{3} x^{6-3} \left( -\frac{3}{2x} \right)^3\)
\(= \binom{6}{3} x^3 \left( -\frac{3}{2x} \right)^3\)
\(= 20 \cdot x^3 \cdot \left( -\frac{3}{2x} \right)^3\)
\(= 20 \cdot x^3 \cdot \left( -\frac{27}{8x^3} \right)\)
\(= 20 \cdot \left( -\frac{27}{8} \right)\)
\(= -67.5\)
