To find the term independent of \(x\) in the expansion of \(\left(x - \frac{1}{x}\right)^4 \left(x + \frac{1}{x}\right)^3\), we need to first understand how the expansion works.
The expression is a product of two binomials, \(\left(x - \frac{1}{x}\right)^4\) and \(\left(x + \frac{1}{x}\right)^3\). Let's expand each binomial separately using the binomial theorem:
\(\left(x - \frac{1}{x}\right)^4 = \sum_{k=0}^{4} \binom{4}{k} x^{4-k} \left(-\frac{1}{x}\right)^k = \sum_{k=0}^{4} \binom{4}{k} (-1)^k x^{4-2k}\)
\(\left(x + \frac{1}{x}\right)^3 = \sum_{j=0}^{3} \binom{3}{j} x^{3-j} \left(\frac{1}{x}\right)^j = \sum_{j=0}^{3} \binom{3}{j} x^{3-2j}\)
We are interested in the term independent of \(x\), which means we want the coefficient of the term where the power of \(x\) is zero after multiplying both expanded forms.
The general term in the expansion is:
\(T = \binom{4}{k} (-1)^k x^{4-2k} \cdot \binom{3}{j} x^{3-2j}\)
For the term to be independent of \(x\), the powers of \(x\) must satisfy:
\(4 - 2k + 3 - 2j = 0\)
Simplifying gives:
\(7 = 2k + 2j\)
\(k + j = \frac{7}{2}\)
Since \(k\) and \(j\) are integers, and \(\frac{7}{2}\) is not an integer, there are no valid integer solutions for \(k\) and \(j\) that satisfy this equation.
Thus, there is no term in the product that is independent of \(x\).
Therefore, the term independent of \(x\) in the expansion is \(0\).