Using binomial theorem, expand each of the following:

Question:

Using binomial theorem, expand each of the following:

$(2 x-3)^{6}$

Solution:

To find: Expansion of $(2 x-3)^{6}$

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, $(2 x-3)^{6}$

$\Rightarrow\left[{ }^{6} \mathrm{C}_{0}(2 \mathrm{x})^{6}\right]+\left[{ }^{6} \mathrm{C}_{1}(2 \mathrm{x})^{6-1}(-3)^{1}\right]+\left[{ }^{6} \mathrm{C}_{2}(2 \mathrm{x})^{6-2}(-3)^{2}\right]+\left[{ }^{6} \mathrm{C}_{3}(2 \mathrm{x})^{6-3}(-3)^{3}\right]+\left[{ }^{6} \mathrm{C}_{4}(2 \mathrm{x})^{6-4}(-3)^{4}\right]+$

$\left[{ }^{6} \mathrm{C}_{5}(2 \mathrm{x})^{6-5}(-3)^{5}\right]+\left[{ }^{6} \mathrm{C}_{6}(-3)^{6}\right]$

$\Rightarrow\left[\frac{6 !}{0 !(6-0) !}(2 x)^{6}\right]-\left[\frac{6 !}{1 !(6-1) !}(2 x)^{5}(3)\right]+\left[\frac{6 !}{2 !(6-2) !}(2 x)^{4}(9)\right]$

$-\left[\frac{6 !}{3 !(6-3) !}(2 x)^{3}(27)\right]+\left[\frac{6 !}{4 !(6-4) !}(2 x)^{2}(81)\right]$

$-\left[\frac{6 !}{5 !(6-5) !}(2 x)^{1}(243)\right]+\left[\frac{6 !}{6 !(6-6) !}(729)\right]$

$\Rightarrow\left[(1)\left(64 x^{6}\right)\right]-\left[(6)\left(32 x^{5}\right)(3)\right]+\left[15\left(16 x^{4}\right)(9)\right]-\left[20\left(8 x^{3}\right)(27)\right]+\left[15\left(4 x^{2}\right)(81)\right]-$

$[(6)(2 x)(243)]+[(1)(729)]$

$\Rightarrow 64 x^{6}-576 x^{5}+2160 x^{4}-4320 x^{3}+4860 x^{2}-2916 x+729$

Ans) $64 x^{6}-576 x^{5}+2160 x^{4}-4320 x^{3}+4860 x^{2}-2916 x+729$