Q68 / 170 · Manipal (MET) 2015 · Mathematics

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
    $-3$
  • Correct answer
    B
    $0$
  • C
    $1$
  • D
    $3$

Solution

We have, $\begin{gathered}\left(x-\frac{1}{x}\right)^4\left(x+\frac{1}{x}\right)^3 \\ =\left({ }^4 C_0 x^4-{ }^4 C_1 x{ }^2+{ }^4 C_2-{ }^4 C_3 \frac{1}{x^2}+{ }^4 C_4 \frac{1}{x^4}\right) \\ \quad \times\left({ }^3 C_0 x^3+{ }^3 C_1 x+{ }^3 C_2 \frac{1}{x}+{ }^3 C_3 \frac{1}{x^3}\right)\end{gathered}$ Clearly, there is no term free from $x$ on RHS. Therefore, the term independent of $x$ on LHS is zero.

Hence, the correct answer is option: 'B'.

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