If $(3+2\sqrt{2}{)}^{{\mathrm{x}}^2-4}+(3-2\sqrt{2}{)}^{{\mathrm{x}}^2-4}=6$, then ${\mathrm{x}}^4+{\mathrm{x}}^2+5=$

$(3+2\sqrt{2}{)}^{{\mathrm{x}}^2-4}+(3-2\sqrt{2}{)}^{{\mathrm{x}}^2-4}=6$

Put,${\left(3+2\sqrt{2}\right)}^{x^2-4}=t$

$t+\frac{1}{t}=6$

$t^2-6t+1=0$

$$\begin{gathered} t=\frac{6\pm 4\sqrt{2}}{2} \\ t=3\pm 2\sqrt{2} \end{gathered}$$

${\left(3+2\sqrt{2}\right)}^{x^2-4}=\left(3+2\sqrt{2}\right)$ and ${\left(3+2\sqrt{2}\right)}^{x^2-4}=\left(3-2\sqrt{2}\right)$

$x^2-4=1$ and $x^2-4=-1$