Degree of the polynomial ${\left[\sqrt{{\mathrm{x}}^2+1}+\sqrt{{\mathrm{x}}^2-1}\right]}^8+{\left[\frac{2}{\sqrt{{\mathrm{x}}^2+1}+\sqrt{{\mathrm{x}}^2-1}}\right]}^8$is ____.

$$\begin{gathered} {\left[\sqrt{{\mathrm{x}}^2+1}+\sqrt{{\mathrm{x}}^2-1}\right]}^8+{\left[\sqrt{{\mathrm{x}}^2+1}-\sqrt{{\mathrm{x}}^2-1}\right]}^8 \\ =2\left[{ }^8{\mathrm{C}}_0{\left(\sqrt{{\mathrm{x}}^2+1}\right)}^8{+}^8{\mathrm{C}}_2{\left(\sqrt{{\mathrm{x}}^2+1}\right)}^6{\left(\sqrt{{\mathrm{x}}^2-1}\right)}^2\right.{+}^8{\mathrm{C}}_4{\left(\sqrt{{\mathrm{x}}^2+1}\right)}^4{\left(\sqrt{{\mathrm{x}}^2-1}\right)}^4 \\ \left.{+}^8{\mathrm{C}}_6{\left(\sqrt{{\mathrm{x}}^2+1}\right)}^2{\left(\sqrt{{\mathrm{x}}^2-1}\right)}^6{+}^8{\mathrm{C}}_8{\left(\sqrt{{\mathrm{x}}^2-1}\right)}^8\right] \end{gathered}$$
which has degree 8.