# Using binomial theorem, expand each of the following:

Question:

Using binomial theorem, expand each of the following:

$\left(\frac{2 x}{3}-\frac{3}{2 x}\right)^{6}$

Solution:

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

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, $\left(\frac{2 x}{3}-\frac{3}{2 x}\right)^{6}$

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

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

$+\left[{ }^{6} C_{5}\left(\frac{2 x}{3}\right)^{6-5}\left(-\frac{3}{2 x}\right)^{5}\right]+\left[{ }^{6} C_{6}\left(-\frac{3}{2 x}\right)^{6}\right]$

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

$\left[\frac{6 !}{2 !(6-2) !}\left(\frac{2 x}{3}\right)^{4}\left(\frac{9}{4 x^{2}}\right)\right]-\left[\frac{6 !}{3 !(6-3) !}\left(\frac{2 x}{3}\right)^{3}\left(\frac{27}{8 x^{3}}\right)\right]+$

$\left[\frac{6 !}{4 !(6-4) !}\left(\frac{2 x}{3}\right)^{2}\left(\frac{81}{16 x^{4}}\right)\right]-\left[\frac{6 !}{5 !(6-5) !}\left(\frac{2 x}{3}\right)^{1}\left(\frac{243}{32 x^{5}}\right)\right]$

$+\left[\frac{6 !}{6 !(6-6) !}\left(\frac{729}{64 x^{6}}\right)\right]$

$\Rightarrow\left[1\left(\frac{64 x^{6}}{729}\right)\right]-\left[6\left(\frac{32 x^{5}}{243}\right)\left(\frac{3}{2 x}\right)\right]+\left[15\left(\frac{16 x^{4}}{81}\right)\left(\frac{9}{4 x^{2}}\right)\right]-\left[20\left(\frac{8 x^{3}}{27}\right)\right.$

$\left.\left(\frac{27}{8 x^{3}}\right)\right]+\left[15\left(\frac{4 x^{2}}{9}\right)\left(\frac{81}{16 x^{4}}\right)\right]-\left[6\left(\frac{2 x}{3}\right)\left(\frac{243}{32 x^{5}}\right)\right]+\left[1\left(\frac{729}{64 x^{6}}\right)\right]$

$\Rightarrow \frac{64}{729} x^{6}-\frac{32}{27} x^{4}+\frac{20}{3} x^{2}-20+\frac{135}{4} \frac{1}{x^{2}}-\frac{243}{8} \frac{1}{x^{4}}+\frac{729}{64} \frac{1}{x^{6}}$

Ans) $\frac{64}{729} x^{6}-\frac{32}{27} x^{4}+\frac{20}{3} x^{2}-20+\frac{135}{4} \frac{1}{x^{2}}-\frac{243}{8} \frac{1}{x^{4}}+\frac{729}{64} \frac{1}{x^{6}}$