$6.279 \sqrt[4]{x+8}-\sqrt[4]{x-8}=2$
Solution. Let $u=\sqrt[4]{x+8}, v=\sqrt[4]{x-8}$, then $\left\{\begin{array}{l}u-v=2, \\ u^{4}-v^{4}=16\end{array} \Rightarrow\right.$ $\Rightarrow(v+2)^{4}-v^{4}=16 \Leftrightarrow\left((v+2)^{2}-v^{2}\right)\left((v+2)^{2}+v^{2}\right)=16 \Leftrightarrow(v+1)(v+2 v+2)=2$.

Since $v \geq 0$, then $v+1 \geq 1, v^{2}+2 v+2 \geq 2 \Rightarrow v=0$ is the only solution to the last equation.

Answer: $x=8$.