CBSE BoardEnglish MediumSTD 11 ScienceMathsModel Paper 63 Marks
Question
Using binomial theorem, expand: $(\sqrt[3]{x}-\sqrt[3]{y})^6$
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Answer
To find: Expension of $(\sqrt[3]{x}-\sqrt[3]{y})^6$ by means of binomial theorem..
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, $(\sqrt[3]{x}-\sqrt[3]{y})^6$
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 $\left(x^{\frac{1}{3}}-y^{\frac{1}{3}}\right)^6$
$\Rightarrow\left[6 C _0\left( x ^{\frac{1}{3}}\right)^{6-0}\right]+\left[6 C _1\left( x ^{\frac{1}{3}}\right)^{6-1}\left(- y ^{\frac{1}{3}}\right)^1\right]+\left[6 C _2\left( x ^{\frac{1}{3}}\right)^{6-2}\left(- y ^{\frac{1}{3}}\right)^2\right]$
$+\left[6 C_3\left(x^{\frac{1}{3}}\right)^{6-3}\left(-y^{\frac{1}{3}}\right)\right] +\left[6 C _4\left( x ^{\frac{1}{3}}\right)^{6-4}\left(- y ^{\frac{1}{3}}\right)^4\right]$
$+\left[6 C _5\left( x ^{\frac{1}{3}}\right)^{6-5}\left(- y ^{\frac{1}{3}}\right)^5\right]+\left[6 C_6\left(-y^{\frac{1}{3}}\right)^6\right]$
$\Rightarrow\left[{ }^6 C _0\left(\frac{6}{x^3}\right)\right]-\left[{ }^6 C _1\left(x^{\frac{5}{3}}\right)\left(y^{\frac{1}{3}}\right)\right]+\left[6 C _2\left( x ^{\frac{4}{3}}\right)\left( y ^{\frac{2}{3}}\right)\right]$
$-\left[6 C _3\left( x ^{\frac{3}{3}}\right)\left( y ^{\frac{3}{3}}\right)\right]+\left[{ }_6 C _4\left( x ^{\frac{2}{3}}\right)\left( y ^{\frac{4}{3}}\right)\right]-\left[6 C _5\left( x ^{\frac{1}{3}}\right)\left( y ^{\frac{5}{3}}\right)\right]$
$+\left[6 C_6\left(\frac{6}{y^3}\right)\right]$
$\Rightarrow\left[\frac{6!}{0!(6-0)!}\left(x^2\right)\right]-\left[\frac{6!}{1!(6-1)!}\left(x^{\frac{5}{3}}\right)\left(y^{\frac{2}{3}}\right)\right]+\left[\frac{6!}{2!(6-2)!}\left(x^{\frac{4}{2}}\right)\left(x^{\frac{2}{3}}\right)\right]$
$-\left[\frac{6!}{3!(6-3)!}(x)(y)\right]+\left[\frac{6!}{4!(6-4)!}\left(x^{\frac{2}{3}}\right)\left(y^{\frac{4}{3}}\right)\right]$
$-\left[\frac{6!}{5!(6-5)!}\left(x^{\frac{1}{3}}\right)\left(y^{\frac{5}{3}}\right)\right]+\left[\frac{6!}{6!(6-6)!}\left(y^2\right)\right]$
$\Rightarrow\left[1\left(x^2\right)\right]-\left[6\left(x^{\frac{5}{3}}\right)\left(y^{\frac{1}{3}}\right)\right]+\left[15\left(x^{\frac{4}{3}}\right)\left(y^{\frac{2}{3}}\right)\right]$
$-[20(x)(y)]+\left[15\left(x^{\frac{2}{3}}\right)\left(\frac{4}{y^3}\right)\right]$
$-\left[6\left(x^{\frac{1}{3}}\right)\left(y^{\frac{5}{3}}\right)\right]+\left[1\left(y^2\right)\right]$
$\Rightarrow x^2-6 x^{\frac{5}{2}} y^{\frac{1}{3}}+15 x^{\frac{4}{3}} y^{\frac{2}{3}} -20 x y+15 x^{\frac{2}{3}} y^{\frac{4}{3}}-6 x^{\frac{1}{3}} y^{\frac{5}{3}}+y^2$
Hence the result.
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