Question

The coefficient of 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. $1$
• D. $3$

Answer 1

The correct option is B $0$

$(x-\frac{1}{x}{)}^4(x+\frac{1}{x}{)}^3$
$=(x-\frac{1}{x})(x^2-\frac{1}{x^2}{)}^3$
$=x(x^2-\frac{1}{x^2}{)}^3-\frac{1}{x}((x^2-\frac{1}{x^2}{)}^3)$
$T_{r+1}=3C_r{x}^{8-5r}$ and $T_{r+1}^{{}^{\prime }}=3C_{r^{\prime }}{x}^{4-5r^{\prime }}$
For term independent of x
$8-5r=0$
$r=\frac{8}{5}$
and $4-5r^{\prime }=0$
$r^{\prime }=\frac{4}{5}$
However, $r$ and $r^{\prime }$ has to be whole numbers, but in the above question, their values come out to be fraction.
That is contradictory.
Hence there would be no term independent of $x$.