4. In the set of real numbers, solve the equation:

$$
\sqrt[3]{x+2}+\sqrt[3]{3 x+1}=\sqrt[3]{x-3}
$$
4. By raising both sides of the equation to the third power (using the identity $\left.(a+b)^{3}=a^{3}+3 a b(a+b)+b^{3}\right)$, we get

$$
x+2+3 \sqrt[3]{x+2} \sqrt[3]{3 x+1}(\sqrt[3]{x+2}+\sqrt[3]{3 x+1})+3 x+1=x-3
$$

i.e.,

$$
3 \sqrt[3]{x+2} \sqrt[3]{3 x+1}(\sqrt[3]{x+2}+\sqrt[3]{3 x+1})=-3 x-6
$$

The expression in parentheses represents the left side of the original equation again, so it must be equal to $\sqrt[3]{x-3}$; by substituting this (and dividing both sides by 3) we arrive at

$$
\sqrt[3]{x+2} \sqrt[3]{3 x+1} \sqrt[3]{x-3}=-(x+2)
$$

Note, the last equation is not equivalent to the original (it is only a consequence of it); this means that for any solution found to the last equation, we must check whether it actually satisfies the original equation.

By raising both sides of the last equation to the third power, we get

$$
(x+2)(3 x+1)(x-3)=-(x+2)^{3}
$$

One solution is clearly $x=-2$. It is directly verified that this is indeed a solution to the original equation (both sides equal $\sqrt[3]{-5}$). Assuming $x \neq-2$, by canceling $x+2$ from both sides, we are left with $(3 x+1)(x-3)=-(x+2)^{2}$, i.e., $3 x^{2}-9 x+x-3=-\left(x^{2}+4 x+4\right)$, which is equivalent to $4 x^{2}-4 x+1=0$. Solving this quadratic equation, we get $x=\frac{4 \pm \sqrt{(-4)^{2}-4 \cdot 4}}{8}=\frac{4 \pm 0}{8}=\frac{1}{2}$. Let's check if this solution satisfies the original equation. On the left side, we get $\sqrt[3]{\frac{1}{2}+2}+\sqrt[3]{3 \cdot \frac{1}{2}+1}=2 \sqrt[3]{\frac{5}{2}}$, and on the right side, $\sqrt[3]{\frac{1}{2}-3}=\sqrt[3]{-\frac{5}{2}}$, so $x=\frac{1}{2}$ is not a solution to the original equation.

Therefore, the only solution is $x=-2$.