**Question:**

Show that the coefficient of $\mathbf{x}^{-3}$ in the expansion of $\left(x-\frac{1}{x}\right)^{11}$ is $-330$.

**Solution:**

To show: that the coefficient of $x^{-3}$ in the expansion of $\left(x-\frac{1}{x}\right)^{11}$ is -330.

Formula Used:

General term, $T_{r+1}$ of binomial expansion $(x+y)^{n}$ is is given by,

$T_{r+1}={ }^{n} C_{r} x^{n-r} y^{r}$ where

${ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}=\frac{n !}{r !(n-r) !}$

Now, finding the general term of the expression, $\left(x-\frac{1}{x}\right)^{11}$, we get

$T_{r+1}={ }^{11} C_{r} \times x^{11-r} \times\left(\frac{-1}{x}\right)^{r}$

For finding the term which has ${ }^{x^{-3}}$ in it, is given by

$11-2 r=3$

$2 r=14$

$R=7$

Thus, the term which the term which has ${ }^{x^{-3}}$ in it is $T_{8}$

$T_{8}={ }^{11} C_{7} \times x^{11-7} \times\left(\frac{-1}{x}\right)^{7}$

$T_{8}=-{ }^{11} C_{7} \times x^{-3}$

$T_{8}=-\frac{11 !}{7 !(11-7) !}$

$\mathrm{T}_{6}=-\frac{11 \times 10 \times 9 \times 8 \times 7 !}{7 ! \times 4 \times 3 \times 2}$

$T_{6}=-330$

Thus, the coefficient of $x^{-3}$ in the expansion of $\left(x-\frac{1}{x}\right)^{11}$ is $-330$.