Example 20. Solve the equation

$$
\sqrt{x^{2}(x-1)}=|x|
$$
Solution.

The first method. The domain of definition (DOD) of equation (8) is given by the condition $x^{2}(x-1) \geqslant 0$, i.e., $x=0$ and $x \geqslant 1$. Let's divide the DOD into two parts: $x=0$ and $x \geqslant 1$.

If $x=0$, then equation (8) turns into a true numerical equality. Thus, the number 0 is a solution to equation (8).

If $x \geqslant 1$, then $|x|=x$ and $\sqrt{x^{2}(x-1)}=x \sqrt{x-1}$; therefore, equation (8) takes the form $x \sqrt{x-1}=x$. Under these conditions, the left and right sides of the last equation can be divided by $x$. Then we have the equation $\sqrt{x-1}=1$, which is equivalent to the equation $x=2$.

Combining the solutions of equation (8) found on both parts of its DOD, we get $x_{1}=0$ and $x_{2}=2$ - the solutions of equation (8).

The second method. The equation $\sqrt{\alpha(x)}=\beta(x)$ is equivalent to the system

$$
\left\{\begin{array}{l}
\beta(x) \geqslant 0 \\
\alpha(x)=\beta^{2}(x)
\end{array}\right.
$$

By definition, the absolute value of a number is a non-negative quantity, therefore,

$$
\text { (8) } \Leftrightarrow x^{2}(x-1)=x^{2} \Leftrightarrow x^{2}(x-2)=0 \Leftrightarrow\left[\begin{array}{l}
x=0 \\
x=2
\end{array}\right.
$$

The third method. Since

$$
f(x)=g(x) \Rightarrow f^{2}(x)=g^{2}(x)
$$

we have

$$
\text { (8) } \Rightarrow x^{2}(x-1)=x^{2} \Leftrightarrow x^{2}(x-2)=0 \Leftrightarrow\left[\begin{array}{l}
x=0 \\
x=2
\end{array}\right.
$$

In solving the equation, a transition to a consequence was used, so it is necessary to perform a check. Substituting $x=0$ and $x=2$ into the original equation, we get true numerical equalities; therefore, $x=0$ and $x=2$ are the roots of equation (8).