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$(\sqrt 3+1)^4+(\sqrt 3-1)^4$ is equal to

$\begin{array}{1 1}(A)\;\text{a rational number}\\(B)\;\text{an irrational number}\\(C)\;\text{a negative integer}\\(D)\;\text{none of these}\end{array} $

1 Answer

$(\sqrt 3+1)^4+(\sqrt 3-1)^4$
$\Rightarrow [(\sqrt 3)^4+4(\sqrt 3)^3+6(\sqrt 3)^2+4(\sqrt 3)+1]+[(\sqrt 3)^4-(\sqrt 3)^3+6(\sqrt 3)^2-4(\sqrt 3)+1]$
$\Rightarrow 2[(\sqrt 3)^4+6(\sqrt 3)^2+1]$
$\Rightarrow 2(9+18+1)=56$
Hence(A) is the correct answer.
answered Jun 25, 2014 by sreemathi.v

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