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Home  >>  CBSE XI  >>  Math  >>  Binomial Theorem
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Evaluate $(\sqrt 3+2)^6-(\sqrt 3-\sqrt 2)^6$

$\begin{array}{1 1}(A)\;396\sqrt 6\\(B)\;396\\(C)\;396\sqrt 2\\(D)\;396\sqrt 3\end{array} $

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1 Answer

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Toolbox:
  • $(a+b)^n=nC_0a^n+nC_1a^{n-1}b+nC_2a^{n-2}b^2+......+nC_ra^{n-r}b^r+.....nC_nb^n$
  • $(a-b)^n=nC_0a^n-nC_1a^{n-1}b+nC_2a^{n-2}b^2-......+(-1)^nnC_ra^{n-r}b^r+.....+nC_n(-b)^n$
$(\sqrt 3+\sqrt 2)^6=(\sqrt 3)^6+6C_1(\sqrt 3)^5(\sqrt 2)+6C_2(\sqrt 3)^4(\sqrt 2)+6C_3(\sqrt 3)^3(\sqrt 2)^3+6C_4(\sqrt 3)^2(\sqrt 2)^4+6C_5(\sqrt 3)(\sqrt 2)^5+(\sqrt 2)^5.......(1)$
$(\sqrt 3-\sqrt 2)^6=(\sqrt 3)^6-6C_1(\sqrt 3)^5(\sqrt 2)+6C_2(\sqrt 3)^4(\sqrt 2)-6C_3(\sqrt 3)^3(\sqrt 2)^3+6C_4(\sqrt 3)^2(\sqrt 2)^4-6C_5(\sqrt 3)(\sqrt 2)^5+(\sqrt 2)^5.......(2)$
Subtracting (2) from (1)
$(\sqrt 3+2)^6-(\sqrt 3-\sqrt 2)^6=2[6C_1(\sqrt 3)^5(\sqrt 2)+6C_3(\sqrt 3)^3(\sqrt 2)^3+6C_5(\sqrt 3)(\sqrt 2)^5]$
$\Rightarrow 2[6\times 3^{5/2}\times 2^{1/2}+20\times3^{3/2}2^{3/2}+6\times3^{1/2}2^{5/2}]$
$\Rightarrow 2\times3^{1/2}\times 2^{1/2}[6.3^2+20\times 3^2+20\times 3\times 2+6\times 2^2]$
$\Rightarrow 2\sqrt 6[54+120+24]$
$\Rightarrow 2\sqrt 6\times 198$
$\Rightarrow 396\sqrt 6$
Hence (A) is the correct answer.
answered Jun 24, 2014 by sreemathi.v
 

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