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Q.
The term independent of $x$ in the expansion of $\left(x-\frac{1}{x}\right)^4\left(x+\frac{1}{x}\right)^3$ is :
Binomial Theorem
Solution:
$\left({ }^4 C _0 x ^4-{ }^4 C _1 x ^2+{ }^4 C _2-4 C _3 \frac{1}{ x ^2}+{ }^4 C _4 \frac{1}{ x ^4}\right)\left( x ^3+3 x +\frac{3}{ x }+\frac{1}{ x ^3}\right) \Rightarrow$ coefficient of $x$ is 0