# Evaluate $(\sqrt 3+\sqrt 2)^4-(\sqrt 3-\sqrt 2)^4$

$\begin{array}{1 1}(A)\;40\\(B)\;\sqrt 6\\(C)\;40\sqrt 6\\(D)\;40\end{array}$

We have $(a+b)^4-(a-b)^4=2[4a^3b+4ab^3]$
$\Rightarrow 8ab(a^2+b^2)$
Put $a=\sqrt 3,b=\sqrt 2$
$\therefore(\sqrt 3+\sqrt 2)^4-(\sqrt 3-\sqrt 2)^4=8.\sqrt 3\sqrt 2[(\sqrt 3)^2+(\sqrt 2^2]$
$\Rightarrow 8\sqrt 6(3+2)$
$\Rightarrow 40\sqrt 6$
Hence (C) is the correct answer.