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

The correct option is B 0

We have,
${(x-\frac{1}{x})}^4{(x+\frac{1}{x})}^3$
$=\left({}^4{C}_0^4{-}^4{C}_1{x}^2{+}^4{C}_2{-}^4{C}_2{-}^4{C}_3\frac{1}{x^2}{+}^4{C}_4\frac{1}{x^4}\right)\times \left({}^3{C}_0{x}^3{+}^3{C}_1{+}^3{C}_2\frac{1}{x}{+}^3{C}_3\frac{1}{x^3}\right)$
Clearly, there is no term free from $x$ on RHS.
Therefore, the term independent of $x$ on LHS is zero.