Using binomial theorem, expand each of the following:

To find: Expansion of $\left(x-\frac{1}{y}\right)^{5}$

Formula used: (i) ${ }^{n} C_{r}=\frac{n !}{(n-r) !(r) !}$

(ii) $(a+b)^{n}={ }^{n} C_{0} a^{n}+{ }^{n} C_{1} a^{n-1} b+{ }^{n} C_{2} a^{n-2} b^{2}+\ldots \ldots+{ }^{n} C_{n-1} a b^{n-1}+{ }^{n} C_{n} b^{n}$

We have, $\left(x-\frac{1}{y}\right)^{5}$

$\Rightarrow{ }^{5} C_{0}(x)^{5-0}+{ }^{5} C_{1}(x)^{5-1}\left(-\frac{1}{y}\right)^{1}+{ }^{5} C_{2}(x)^{5-2}\left(-\frac{1}{y}\right)^{2}+{ }^{5} C_{3}(x)^{5-3}\left(-\frac{1}{y}\right)^{3}+{ }^{5} C_{4}(x)^{5-4}\left(-\frac{1}{y}\right)^{4}+{ }^{5} C_{5}$

$\Rightarrow\left[\frac{5 !}{0 !(5-0) !}\left(x^{5}\right)\right]-\left[\frac{5 !}{1 !(5-1) !}\left(x^{4}\right)\left(\frac{1}{y}\right)\right]+\left[\frac{5 !}{2 !(5-2) !}\left(x^{3}\right)\left(\frac{1}{y^{2}}\right)\right]$

$-\left[\frac{5 !}{3 !(5-3) !}\left(x^{2}\right)\left(\frac{1}{y^{3}}\right)\right]+\left[\frac{5 !}{4 !(5-4) !}(x)\left(\frac{1}{y^{4}}\right)\right]-\left[\frac{5 !}{5 !(5-5) !}\left(\frac{1}{y^{5}}\right)\right]$

⇒ $\left[1\left(x^{5}\right)\right]-\left[5\left(\frac{x^{4}}{y}\right)\right]+\left[10\left(\frac{x^{3}}{y^{2}}\right)\right]-\left[10\left(\frac{x^{2}}{y^{3}}\right)\right]+\left[5\left(\frac{x}{y^{4}}\right)\right]-\left[1\left(y^{5}\right)\right]$

$\Rightarrow x^{5}-5 \frac{x^{4}}{y}+10 \frac{x^{3}}{y^{2}}-10 \frac{x^{2}}{y^{3}}+5 \frac{x}{y^{4}}-y^{5}$

Ans) $x^{5}-5 \frac{x^{4}}{y}+10 \frac{x^{3}}{y^{2}}-10 \frac{x^{2}}{y^{3}}+5 \frac{x}{y^{4}}-y^{5}$