6.043. $\sqrt{x^{2}+32}-2 \sqrt[4]{x^{2}+32}=3$.
## Solution.

Let $\sqrt[4]{x^{2}+32}=y>0$. With respect to $y$, we obtain the equation $y^{2}-2 y-3=0$, from which $y_{1}=-1$ (not suitable), $y_{2}=3$. Then

$$
\sqrt[4]{x^{2}+32}=3 \Leftrightarrow x^{2}+32=81 \Leftrightarrow x^{2}=49
$$

This expression satisfies the given equation. Hence, $|x|=7$ or $x_{1,2}= \pm 7$.

Answer: $x_{1}=-7, x_{2}=7$.