## Solution.

Let's rewrite the equation as $\sqrt[3]{5 x+7}=\sqrt[3]{5 x-12}+1$ and cube both sides:

$$\begin{aligned} & 5 x+7=5 x-12+3(\sqrt[3]{5 x-12})^{2}+3 \sqrt[3]{5 x-12}+1 \Leftrightarrow \\ & \Leftrightarrow(\sqrt[3]{5 x-12})^{2}+\sqrt[3]{5 x-12}-6=0 \end{aligned}$$

Let $\sqrt[3]{5 x-12}=t$. The equation in terms of $t$ becomes $t^{2}-t-6=0$, from which we find $t_{1}=-3$ and $t_{2}=2$.

Then either $\sqrt[3]{5 x-12}=-3, 5 x-12=-27, x_{1}=-3$, or $\sqrt[3]{5 x-12}=2$, $5 x-12=8, x_{2}=4$.

Answer: $x_{1}=-3, x_{2}=4$.