Explanation

(b)
$\left( x - \frac{1}{x} \right)^{4}\left( x + \frac{1}{x} \right)^{3}$
$= \left( \ ^{4}C_{0}x^{4} - \ ^{4}C_{1}x^{2} + \ ^{4}C_{2} - \ ^{4}C_{3}\ \frac{1}{x^{2}} + \ ^{4}C_{4}\frac{1}{x^{4}} \right)$
×$\left( \ ^{3}C_{0}x^{3} + \ ^{3}C_{1}x + \ ^{3}C_{2}\frac{1}{x} + \ ^{3}C_{3}\frac{1}{x^{3}} \right)$
Clearly, there is no term from $x$ on RHS, therefore the term independent of $x$ on LHS is zero.