3. The real number $x$ satisfies
$$
\sqrt{x^{2}+3 \sqrt{2} x+2}-\sqrt{x^{2}-\sqrt{2} x+2}=2 \sqrt{2} x \text {. }
$$

Then the value of the algebraic expression $x^{4}+x^{-4}$ is $\qquad$
3.14.

Obviously $x \neq 0$.
Let $\sqrt{x^{2}+3 \sqrt{2} x+2}=a$,
$$
\sqrt{x^{2}-\sqrt{2} x+2}=b \text {. }
$$

Then $a^{2}-b^{2}=(a+b)(a-b)=4 \sqrt{2} x$.
Also, $a-b=2 \sqrt{2} x$, so, $a+b=2$.
Thus, $a=\sqrt{2} x+1$.
Then $(\sqrt{2} x+1)^{2}=x^{2}+3 \sqrt{2} x+2$.
Rearranging gives $x^{2}-\sqrt{2} x-1=0$, which is $x-x^{-1}=\sqrt{2}$.
Squaring both sides gives $x^{2}-2+x^{-2}=2$, which is
$$
x^{2}+x^{-2}=4 \text {. }
$$

Squaring both sides of the above equation gives $x^{4}+2+x^{-4}=16$, which is $x^{4}+x^{-4}=14$.