Using binomial theorem, expand each of the following:

Solution:

To find: Expansion of $(\sqrt{x}+\sqrt{y})^{8}$

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, $(\sqrt{x}+\sqrt{y})^{8}$

We can write $\sqrt{x}$ as $x^{\frac{1}{2}}$ and $\sqrt{y}$ as $y^{\frac{1}{2}}$

Now, we have to solve for $\left(x^{\frac{1}{2}}+y^{\frac{1}{2}}\right)^{8}$

$\Rightarrow\left[{ }^{8} C_{0}\left(\frac{1}{x^{2}}\right)^{8-0}\right]+\left[{ }^{8} C_{1}\left(x^{\frac{1}{x^{2}}}\right)^{8-1}\left(\frac{1}{y^{2}}\right)^{1}\right]+\left[{ }^{8} C_{2}\left(\frac{1}{x^{2}}\right)^{8-2}\left(\frac{1}{y^{2}}\right)^{2}\right]+$

$\left[{ }^{8} C_{3}\left(\frac{1}{x^{2}}\right)^{8-3}\left(\frac{1}{y^{2}}\right)^{3}\right]+\left[{ }^{8} C_{4}\left(\frac{1}{x^{2}}\right)^{8-4}\left(\frac{1}{y^{2}}\right)^{4}\right]+\left[{ }^{8} C_{5}\left(x^{\frac{1}{2}}\right)^{8-5}\left(\frac{1}{y^{2}}\right)^{5}\right]+$

$\left[{ }^{8} C_{6}\left(x^{\frac{1}{x^{2}}}\right)^{8-6}\left(\frac{1}{y^{2}}\right)^{6}\right]+\left[{ }^{8} C_{7}\left(x^{\frac{1}{2}}\right)^{8-7}\left({y^{2}}^{\frac{1}{2}}\right)^{7}\right]+\left[{ }^{8} C_{8}\left({y^{\frac{1}{2}}}^{8}\right)^{8}\right]$

$\Rightarrow\left[\frac{8 !}{0 !(8-0) !}\left(x^{\frac{8}{2}}\right)\right]+\left[\frac{8 !}{1 !(8-1) !}\left(x^{\frac{7}{2}}\right)\left(\frac{1}{y^{2}}\right)\right]+\left[\frac{8 !}{2 !(8-2) !}\left(x^{\frac{6}{2}}\right)\left(\frac{2}{y^{2}}\right)\right]+$

$\left[\frac{8 !}{3 !(8-3) !}\left(x^{\frac{5}{2}}\right)\left(y^{\frac{3}{2}}\right)\right]+\left[\frac{8 !}{4 !(8-4) !}\left(x^{\frac{4}{2}}\right)\left(\frac{4}{y^{2}}\right)\right]+\left[\frac{8 !}{5 !(8-5) !}\left(x^{\frac{3}{2}}\right)\left(y^{\frac{5}{2}}\right)\right]+$

$\left[\frac{8 !}{6 !(8-6) !}\left(x^{\frac{2}{2}}\right)\left(\frac{6}{y^{2}}\right)\right]+\left[\frac{8 !}{7 !(8-7) !}\left(x^{\frac{1}{2}}\right)\left(y^{\frac{7}{2}}\right)\right]+\left[\frac{8 !}{8 !(8-8) !}\left(y^{\frac{8}{2}}\right)\right]$

$\Rightarrow\left[1\left(x^{4}\right)\right]+\left[8\left(x^{\frac{7}{2}}\right)\left(y^{\frac{1}{2}}\right)\right]+\left[28\left(x^{3}\right)(y)\right]+\left[56\left(x^{\frac{5}{2}}\right)\left(y^{\frac{3}{2}}\right)\right]$

$+\left[70\left(x^{2}\right)\left(y^{2}\right)\right]+\left[56\left(x^{\frac{3}{2}}\right)\left(y^{\frac{5}{2}}\right)\right]+\left[28\left(x^{1}\right)\left(y^{3}\right)\right]+\left[8\left(x^{\frac{1}{2}}\right)\left(y^{\frac{7}{2}}\right)\right]+\left[1\left(y^{4}\right)\right]$

Ans) $\left(x^{4}\right)+8\left(x^{7 / 2}\right)\left(y^{1 / 2}\right)+28\left(x^{3}\right)(y)+56\left(x^{5 / 2}\right)\left(y^{3 / 2}\right)+70\left(x^{2}\right)\left(y^{2}\right)+56^{\left(x^{3 / 2}\right)\left(y^{5 / 2}\right)}+$ $28(x)^{1}(y)^{3}+8^{\left(x^{1 / 2}\right)\left(y^{7 / 2}\right)}+(y)^{4}$