1. Given $a^{2 x}=\sqrt{2}+1(a>0)$, then the value of $\frac{a^{3 x}+a^{3 x}}{a^{x}+a^{x}}$ is $\qquad$ .
一、1.2 $\sqrt{2}-1$ (Hint: Apply the sum of cubes formula)

 untranslated part: 
一、1.2 $\sqrt{2}-1$ (提示:应用立方和公式)

The translated part:
1.2 $\sqrt{2}-1$ (Hint: Apply the sum of cubes formula)