$2.18 \frac{\sqrt[3]{x^{4}}-1}{\sqrt[3]{x^{2}}-1}-\frac{\sqrt[3]{x^{2}}-1}{\sqrt[3]{x}+1}=4$.
2.18 Reducing the fractions in the left part of the equation, we get

$\sqrt[3]{x^{2}}+1-\sqrt[3]{x}+1=4$, or $\sqrt[3]{x^{2}}-\sqrt[3]{x}-2=0$.

From this, $\sqrt[3]{x}=-1$ or $\sqrt[3]{x}=2$. Since $\sqrt[3]{x}$ cannot be equal to -1 (as this would make the denominators of the original fractions zero), then $x=8$.

Omeem: $x=8$.