A) -3
B) 0
C) 3
D) 1
Correct Answer: B
Solution :
\[{{\left( x-\frac{1}{x} \right)}^{4}}{{\left( x+\frac{1}{x} \right)}^{3}}\] \[=\left( {{\,}^{4}}{{C}_{0}}{{x}^{4}}{{-}^{4}}\,{{C}_{1}}{{x}^{2}}{{+}^{4}}\,{{C}_{2}}{{-}^{4}}\,{{C}_{3}}\frac{1}{{{x}^{2}}}{{+}^{4}}{{C}_{4}}\frac{1}{{{x}^{4}}} \right)\times \] \[\left( ^{3}{{C}_{0}}{{x}^{3}}{{+}^{3}}{{C}_{1}}x+{{\,}^{3}}{{C}_{2}}\frac{1}{x}+{{\,}^{3}}{{C}_{3}}\frac{1}{{{x}^{3}}} \right)\] It is clear that there is no term free from x on RHS. Hence, the term independent of x is zero.
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