Find the value of.

Firstly, the expression $(x+y)^{4}+(x-y)^{4}$ is simplified by using Binomial Theorem.

This can be done as

$(x+y)^{4}={ }^{4} C_{0} x^{4}+{ }^{+} C_{1} x^{3} y+{ }^{4} C_{2} x^{2} y^{2}+{ }^{4} C_{3} x y^{3}+{ }^{4} C_{4} y^{4}$

$=x^{4}+4 x^{3} y+6 x^{2} y^{2}+4 x y^{3}+y^{4}$

$(x-y)^{4}={ }^{4} C_{0} x^{4}-{ }^{4} C_{1} x^{5} y+{ }^{4} C_{2} x^{2} y^{2}-{ }^{4} C_{3} x y^{3}+{ }^{4} C_{4} y^{4}$

$=x^{4}-4 x^{3} y+6 x^{2} y^{2}-4 x y^{3}+y^{4}$

$\therefore(x+y)^{4}+(x-y)^{4}=2\left(x^{4}+6 x^{2} y^{2}+y^{4}\right)$

Putting $x=a^{2}$ and $y=\sqrt{a^{2}}-1$, we obtain

$\left(a^{2}+\sqrt{a^{2}-1}\right)^{4}+\left(a^{2}-\sqrt{a^{2}-1}\right)^{4}=2\left[\left(a^{2}\right)^{4}+6\left(a^{2}\right)^{2}\left(\sqrt{a^{2}-1}\right)^{2}+\left(\sqrt{a^{2}-1}\right)^{4}\right]$

$=2\left[a^{8}+6 a^{4}\left(a^{2}-1\right)+\left(a^{2}-1\right)^{2}\right]$

$=2\left[a^{8}+6 a^{6}-6 a^{4}+a^{4}-2 a^{2}+1\right]$

$=2\left[a^{8}+6 a^{6}-5 a^{4}-2 a^{2}+1\right]$

$=2 a^{8}+12 a^{6}-10 a^{4}-4 a^{2}+2$