Using binomial theorem, expand each of the following:

To find: Expansion of $(2 x-3 y)^{4}$

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, $(2 x-3 y)^{4}$

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

$\left[\frac{4 !}{0 !(4-0) !}(2 x)^{4}\right]-\left[\frac{4 !}{1 !(4-1) !}(2 x)^{3}(3 y)\right]+\left[\frac{4 !}{2 !(4-2) !}(2 x)^{2}\left(9 y^{2}\right)\right]-$

$\left[\frac{4 !}{3 !(4-3) !}(2 x)^{1}\left(27 y^{3}\right)\right]+\left[\frac{4 !}{4 !(4-4) !}\left(81 y^{4}\right)\right]$

$\Rightarrow\left[1\left(16 x^{4}\right)\right]-\left[4\left(8 x^{3}\right)(3 y)\right]+\left[6\left(4 x^{2}\right)\left(9 y^{2}\right)\right]-\left[4(2 x)\left(27 y^{3}\right)\right]+\left[1\left(81 y^{4}\right)\right]$

$\Rightarrow 16 x^{4}-96 x^{3} y+216 x^{2} y^{2}-216 x y^{3}+81 y^{4}$

Ans) $16 x^{4}-96 x^{3} y+216 x^{2} y^{2}-216 x y^{3}+81 y^{4}$