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

This can be done as

${(x + y)^4} = {\,^4}{C_0}{x^4} + {\,^4}{C_1}{x^3}y + {\,^4}{C_2}{x^2}{y^2} + {\,^4}{C_3}x{y^3} + {\,^4}{C_4}{y^4}$

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

${(x – y)^4} = {\,^4}{C_0}{x^4} – {\,^4}{C_1}{x^3}y + {\,^4}{C_2}{x^2}{y^2} – {\,^4}{C_3}x{y^3} + {\,^4}{C_4}{y^4}$

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

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

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

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

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

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

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

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