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. $1$
• D. $3$

Answer 1

Answer: (B)

$0$

${\left(x-\frac{1}{x}\right)}^4{\left(x+\frac{1}{x}\right)}^3$

$=\left(x-\frac{1}{x}\right){\left[\left(x-\frac{1}{x}\right)\left(x+\frac{1}{x}\right)\right]}^3$

$=\left(x-\frac{1}{x}\right){\left[x^2-\frac{1}{x^2}\right]}^3$ $\left[\right(a-b\left)\right(a+b)=a^2-b^2]$

$=\left(x-\frac{1}{x}\right)\left[x^6-\frac{1}{x^6}-3x^4\times \frac{1}{x^2}+3x^2\times \frac{1}{x^4}\right]$ $\left[{\left(a-b\right)}^3=a^3-b^3-3a^{2}b+3a{b}^2\right]$

$=\left(x-\frac{1}{x}\right)\left[x^6-\frac{1}{x^6}-3x^2+\frac{3}{x^2}\right]$

$=x^7-\frac{1}{x^5}-3x^3+\frac{3}{x}-x^5+\frac{1}{x^7}+3x-\frac{3}{x^3}$

$=x^7-x^{-5}-3x^3+3x^{-1}-x^5+x^{-7}+3x-3x^{-3}$

Since there is no term which is imdependent of $x$

Therefore term independent of $x$ is $0$