Question

The term independent of $x$ in the expansion of ${(x-\frac{1}{x})}^4{(x+\frac{1}{x})}^3$ is

• A. $-3$
• B. $0$
• C. $3$
• D. $1$

Answer 1

The correct option is B $0$

${(x-\frac{1}{x})}^4{(x+\frac{1}{x})}^3$
$=({}^4{C}_0{x}^4-{}^4{C}_1{x}^2+{}^4{C}_2-{}^4{C}_3\frac{1}{x^2}+{}^4{C}_4\frac{1}{x^4})\times ({}^3{C}_0{x}^3+{}^3{C}_{1}x+{}^3{C}_2\frac{1}{x}+{}^3{C}_3\frac{1}{x^3})$
It is clear that there is no term free from $x$ on RHS.
Hence, the term independent of $x$ is zero.