6.053. $\frac{\sqrt[3]{x^{4}}-1}{\sqrt[3]{x^{2}}-1}-\frac{\sqrt[3]{x^{2}}-1}{\sqrt[3]{x}+1}=4$
## Solution.

Domain of definition: $x \neq \pm 1$.

Let $\sqrt[3]{x}=y, y \neq \pm 1$. The equation in terms of $y$ becomes

$$
\begin{aligned}
& \frac{y^{4}-1}{y^{2}-1}-\frac{y^{2}-1}{y+1}=4 \Leftrightarrow \frac{\left(y^{2}-1\right)\left(y^{2}+1\right)}{y^{2}-1}-\frac{(y-1)(y+1)}{y+1}=4 \Leftrightarrow \\
& \Leftrightarrow y^{2}+1-y+1=4 \Leftrightarrow y^{2}-y-2=0
\end{aligned}
$$

from which we find $y_{1}=-1, y_{2}=2$. Then $\sqrt[3]{x}=-1, x_{1}=-1$, or $\sqrt[3]{x}=2$, $x_{2}=8 ; x_{1}=-1$ does not satisfy the domain of definition.

Answer: $x=8$.