# Using binomial theorem, expand each of the following:

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Question:

Using binomial theorem, expand each of the following:

$(\sqrt[3]{x}-\sqrt[3]{y})^{6}$

Solution:

To find: Expansion of $(\sqrt[3]{x}-\sqrt[3]{y})^{6}$

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, $(\sqrt[3]{x}-\sqrt[3]{y})^{6}$

We can write $\sqrt[3]{x}$ as $x^{\frac{1}{3}}$ and $\sqrt[3]{y}$ as $y^{\frac{1}{3}}$

Now, we have to solve for $\left(x^{\frac{1}{3}}-y^{\frac{1}{3}}\right)^{6}$

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

$\left[{ }^{6} C_{3}\left(x^{\frac{1}{3}}\right)^{6-3}\left(-y^{\frac{1}{3}}\right)^{3}\right]+\left[{ }^{6} C_{4}\left(x^{\frac{1}{3}}\right)^{6-4}\left(-y^{\frac{1}{3}}\right)^{4}\right]+\left[{ }^{6} C_{5}\left(x^{\frac{1}{3}}\right)^{6-5}\left(-y^{\frac{1}{3}}\right)^{5}\right]+$

$\left[{ }^{6} C_{6}\left(-y^{\frac{1}{3}}\right)^{6}\right]$

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

$\left[{ }^{6} C_{4}\left(x^{\frac{2}{3}}\right)\left(\frac{4}{y^{3}}\right)\right]-\left[{ }^{6} C_{5}\left(x^{\frac{1}{3}}\right)\left(y^{\frac{5}{3}}\right)\right]+\left[{ }^{6} C_{6}\left(y^{\frac{6}{3}}\right)\right]$

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

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

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

$\Rightarrow\left[1\left(x^{2}\right)\right]-\left[6\left(x^{\frac{5}{3}}\right)\left(y^{\frac{1}{3}}\right)\right]+\left[15\left(x^{\frac{4}{3}}\right)\left(y^{\frac{2}{3}}\right)\right]-[20(x)(y)]+\left[15\left(x^{\frac{2}{3}}\right)\left(\frac{4}{y^{3}}\right)\right]-$

$\left[6\left(x^{\frac{1}{3}}\right)\left(y^{\frac{5}{3}}\right)\right]+\left[1\left(y^{2}\right)\right]$

$\Rightarrow x^{2}-6 x^{\frac{5}{3}} y^{\frac{1}{3}}+15 x^{\frac{4}{3}} y^{\frac{2}{3}}-20 x y+15 x^{\frac{2}{3}} y^{\frac{4}{3}}-6 x^{\frac{1}{3}} y^{\frac{5}{3}}+y^{2}$

Ans) $x^{2}-6 x^{5 / 3} y^{1 / 3}+15 x^{4 / 3} y^{2 / 3}-20 x y+15 x^{2 / 3} y^{4 / 3}-6 x^{1 / 3} y^{5 / 3}+y^{2}$