6.049. $\sqrt{x^{3}+8}+\sqrt[4]{x^{3}+8}=6$.

6.049. $\sqrt{x^{3}+8}+\sqrt[4]{x^{3}+8}=6$.
## Solution.

Domain of definition: $x^{3}+8 \geq 0 \Leftrightarrow x^{3} \geq-8 \Leftrightarrow x \geq-2$.

Let $\sqrt[4]{x^{3}+8}=y, y>0$, and the equation becomes $y^{2}+y=6 \Leftrightarrow$ $\Leftrightarrow y^{2}+y-6=0$, from which $y_{1}=-3, y_{2}=2 ; y_{1}=-3$ is not suitable. Then $\sqrt[4]{x^{3}+8}=2, x^{3}+8=16, x^{3}=8, x=2$.

Answer: $x=2$.