# Find the value of $(a^2+\sqrt{a^2-1})^4+(a^2-\sqrt{a^2-1})^4$

$\begin{array}{1 1}(A)\;2(a^8+6a^6-5a^4-2a^2+1)\\(B)\;4\\(C)\;(a^8+6a^6-5a^4-2a^2+1)\\(D)\;a^5+6a^4-5a^3-2a+1\end{array}$

Put $a^2=x,y=\sqrt{a^2-1}$
$\therefore (x+y)^4=x^4+4C_1x^3y+4C_2x^2y^2+4C_3xy^3+4C_4y^4$------(1)
$(x-y)^4=x^4-4C_1x^3y+4C_2x^2y^2-4C_3xy^3+4C_4y^4$------(2)
$(x+y)^4+(x-y)^4=2[x^4+4C_2x^2y^2+4C_4y^4]$
$\Rightarrow 2[x^4+6x^2y^2+y^4]$
$\therefore (a^2+\sqrt{a^2-1})^4+(a^2-\sqrt{a^2-1})^4=2[(a^2)^4+6(a^2)^2(\sqrt{a^2-1})^2+(\sqrt{a^2-1})^4]$
$\Rightarrow 2[a^8+6a^4(a^2-1)+(a^2-1)^2]$
$\Rightarrow 2[a^8+6a^4(a^2-1)+a^4-2a^2+1]$
$\Rightarrow 2[a^8+6a^6-5a^4-2a^2+1]$
Hence (A) is the correct answer.