394. Solve the equation:

$$
\sqrt[3]{x}+\sqrt[3]{x+19}=5
$$
$\triangle$ Equations of this type have already been encountered in section 11.1, with a different method of solution (see the solution to problem 390). Let's introduce two new variables:

$$
\sqrt[3]{x}=y, \quad \sqrt[3]{x+19}=z
$$

We obtain a system of rational equations:

$$
\left\{\begin{array}{l}
y+z=5 \\
y^{3}=x \\
z^{3}=x+19
\end{array}\right.
$$

Subtract the second equation from the third to eliminate $x$: $z^{3}-y^{3}=19$.

To solve the system of two equations

$$
y+z=5, \quad z^{3}-y^{3}=19
$$

express $z$ from the first equation and substitute this expression into the second equation:

$z=5-y, \quad(5-y)^{3}-y^{3}=19, \quad 125-75 y+15 y^{2}-2 y^{3}=19$, $2 y^{3}-15 y^{2}+75 y-106=0$.

Among the divisors of the free term, the last equation is satisfied by $y=2$. Then the equation reduces to:

$$
(y-2)\left(2 y^{2}-11 y+51\right)=0 .
$$

From this, it is clear that the equation has no other roots.

Therefore,

$$
x=y^{3}=2^{3}=8
$$

Answer: 8.