Using binomial theorem, expand: (x+y)^8 - Vidyadip

Question

Using binomial theorem, expand: $(\sqrt{x}+\sqrt{y})^8$

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Answer

We hand to find value of $(\sqrt{x}+\sqrt{y})^8$
Formula used: ${ }^n C_r=\frac{n!}{(n-r)!(r)!}$
$( a + b )^{ n }={ }^n C_0 a^n+{ }^n C_1 a^{n-1} b+{ }^n C_2 a^{n-2} b^2+\ldots \ldots+{ }^n C_{n-1} a b^{n-1}+{ }^n C_n b^n$
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 $\left(x^{\frac{1}{2}}+y^{\frac{1}{2}}\right)^8$
$\begin{array}{l}=\left[8 C_0\left(x^{\frac{2}{2}}\right)^{8-0}\right]+\left[8 C_1\left(x^{\frac{1}{2}}\right)^{8-1}\left(y^{\frac{2}{2}}\right)^1\right]+\left[8 C_2\left(x^{\frac{1}{2}}\right)^{8-2}\left(y^{\frac{1}{2}}\right)^2\right]+\left[{ }^8 C_3\left(x^{\frac{1}{2}}\right)^{8-3}\left(y^{\frac{1}{2}}\right)^3\right] \\
+\left[{ }^8 C_4\left(x^{\frac{1}{2}}\right)^{8-4}\left(\frac{1}{y^2}\right)^4\right]+\left[{ }^8 C_5\left(x^{\frac{1}{2}}\right)^{8-5}\left(y^{\frac{2}{2}}\right)^5\right]+\left[{ }^8 C_6\left(x^{\frac{1}{2}}\right)^{8-6}\left(y^{\frac{1}{2}}\right)^6\right] \\
+\left[8 C_7\left(x^{\frac{1}{2}}\right)^{8-7}\left(y^{\frac{1}{2}}\right)^7\right]+\left[8 C_8\left(y^{\frac{1}{2}}\right)^8\right] \\
=\left[\frac{81}{0(8-0)!}\left(x^{\frac{5}{2}}\right)\right]+\left[\frac{81}{11(8-1)!}\left(x^{\frac{2}{2}}\right)\left(y^{\frac{1}{2}}\right)\right]+\left[\frac{81}{2(8-2)!}\left(x^{\frac{6}{2}}\right)\left(y^{\frac{2}{2}}\right)\right] \\
+\left[\frac{81}{3(8-3)!}\left(x^{\frac{5}{2}}\right)\left(y^{\frac{3}{2}}\right)\right]+\left[\frac{81}{4(8-4)!}\left(x^{\frac{4}{2}}\right)\left(y^{\frac{4}{2}}\right)\right] \\
+\left[\frac{81}{5(8-5)!}\left(x^{\frac{2}{2}}\right)\left(y^{\frac{5}{2}}\right)\right]+\left[\frac{81}{6(8-6)!}\left(x^{\frac{2}{2}}\right)\left(\frac{6}{y^2}\right)\right]+\left[\frac{81}{7(8-7)!}\left(x^{\frac{1}{2}}\right)\left(\frac{2}{y^2}\right)\right]+\left[\frac{81}{8(8-8)( }\left(y^{\frac{5}{2}}\right)\right] \\
=\left[1\left(x^4\right)\right]+\left[8\left(x^{\frac{7}{2}}\right)\left(y^{\frac{1}{2}}\right)\right]+\left[28\left(x^3\right)(y)\right]+\left[56\left(x^{\frac{5}{2}}\right)\left(\frac{2}{y^2}\right)\right] \\
+\left[70\left(x^2\right)\left(y^2\right)\right]+\left[56\left(x^{\frac{3}{2}}\right)\left(y^{\frac{5}{2}}\right)\right]+\left[28\left(x^2\right)\left(y^3\right)\right]+\left[8\left(x^{\frac{1}{2}}\right)\left(y^{\frac{2}{2}}\right)\right]+\left[1\left(y^4\right)\right]\end{array}$

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