390. Solve the equation:

$$
\sqrt[3]{x-1}+\sqrt[3]{2 x-1}=1
$$
$\triangle$ Let's raise both sides of the equation to the third power, using the formula

$$
(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)
$$

We will have:

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

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

But what now? Now let's use the original equation, according to which the sum in parentheses equals 1:

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

We will also raise the last equation to the third power:

$(x-1)(2 x-1)=(1-x)^{3}, \quad(x-1)\left(2 x-1+(x-1)^{2}\right)=0$, $(x-1) x^{2}=0$.

From this

$$
x_{1}=1, \quad x_{2}=x_{3}=0
$$

It is interesting to check the original equation. It shows that the value $x=1$ satisfies it, while the value $x=0$ does not satisfy it.

Answer: 1.

A problem for reflection: Do you think at which step in the transition from the given original equation to the last one and for what reason did the extraneous root $x=0$ of the original equation appear? After all, it seems that we only used raising the equation to the third power, and as a result, we get an equation equivalent to the original one...

Note. The solution to problem 390 shows that when solving equations of the type

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

using this method, extraneous roots can appear. (When solving such equations by other methods, they may not appear.) Therefore, it is necessary to check the obtained roots by the original equation.