Question

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

• A. $-3$
• B. $0$
• C. $1$
• D. $3$

Answer 1

Answer: (B)

$\left(x-\frac{1}{x}\right){\left(x-\frac{1}{x}\right)}^3{\left(x+\frac{1}{x}\right)}^3$
$=\left(x-\frac{1}{x}\right){\left(x^2-\frac{1}{x^2}\right)}^3$
$=\left(x-\frac{1}{x}\right)\left(x^6-3x^2+3\frac{1}{x^2}-\frac{1}{x^6}\right)$
There is no term containing $\frac{1}{x}$ and $x$ in the second bracket.
So, the expansion will not contain any term independent of $x$ .