Solve the equation

$$
\left(\frac{x}{2}\right)^{2}=\left(\frac{x+1}{3}\right)^{3}
$$
Solution. Let's perform the indicated operations and try to factor the resulting equation:

$$
\begin{gathered}
27 x^{2}=4 x^{3}+12 x^{2}+12 x+4 \\
4 x^{3}-15 x^{2}+12 x+4=0
\end{gathered}
$$

Further transforming:

$$
\begin{aligned}
4 x^{3}-8 x^{2}-7 x^{2}+12 x+4=4 x^{2}(x-2)-7 x^{2}+14 x+4-2 x & = \\
=4 x^{2}(x-2)-7 x(x-2)-2(x-2)=(x-2)\left(4 x^{2}-7 x-2\right) & =0
\end{aligned}
$$

A product is zero if and only if one of its factors is zero. Therefore, either $x-2=0$, and in this case $x=2$, or $4 x^{2}-7 x-2=0$. The solutions of the quadratic equation are:

$$
x_{1,2}=\frac{7 \pm \sqrt{81}}{8}
$$

from which $x_{1}=2$ (which we already obtained earlier) and $x_{2}=-\frac{1}{4}$. Since we performed equivalent transformations throughout, both numbers are solutions to the equation.