Solve the equation

$$
x^{3}+\left(\frac{x}{2 x-1}\right)^{3}=\frac{243}{64}
$$
I. solution. The denominator cannot be zero, so $x \neq \frac{1}{2}$. On the left side of the equation, there is a sum of two cubes, for which we will use the sum of cubes formula. For this purpose, let's first write the cube of the sum of $x$ and $\frac{x}{2 x-1}$ appearing on the left side. On the one hand,

$$
\left(x+\frac{x}{2 x-1}\right)^{3}=\left(\frac{(2 x-1) x+x}{2 x-1}\right)^{3}=\left(\frac{2 x^{2}}{2 x-1}\right)^{3}
$$

on the other hand, according to the cubic identity, using the equation,

$$
\begin{aligned}
\left(x+\frac{x}{2 x-1}\right)^{3} & =x^{3}+\left(\frac{x}{2 x-1}\right)^{3}+3 x^{2} \frac{x}{2 x-1}+3 x\left(\frac{x}{2 x-1}\right)^{2}= \\
& =\frac{243}{64}+3\left(\frac{x^{3}}{2 x-1}+\frac{x^{3}}{(2 x-1)^{2}}\right)= \\
& =\frac{243}{64}+3\left(\frac{(2 x-1) x^{3}+x^{3}}{(2 x-1)^{2}}\right)=\frac{243}{64}+3\left(\frac{2 x^{4}}{(2 x-1)^{2}}\right)
\end{aligned}
$$

By equating the two ways of writing it,

$$
\left(\frac{2 x^{2}}{2 x-1}\right)^{3}=\frac{243}{64}+3\left(\frac{2 x^{4}}{(2 x-1)^{2}}\right)
$$

Multiply both sides of the equation by 64, then introduce the new unknown $y$ for $\frac{8 x^{2}}{2 x-1}$:

$$
64\left(\frac{2 x^{2}}{2 x-1}\right)^{3}=243+3 \cdot 64\left(\frac{2 x^{4}}{(2 x-1)^{2}}\right), \quad y^{3}-6 y^{2}-243=0
$$

If the equation has a rational root, which is expressed in the form $y=\frac{p}{q}$, then $p \mid a_{0}=-243$ and $q \mid a_{3}=1$, so the rational root is an integer. Since $243=3^{5}$, it can be seen after at most five attempts that the equation has the solution $y=9$, because

$$
9^{3}-6 \cdot 81-243=729-486-243=0
$$

Knowing the found root, transform the equation into a product:

$$
(y-9)\left(y^{2}+3 y+27\right)=0
$$

The second factor cannot be zero because the discriminant of the equation $y^{2}+3 y+27=0$ is negative. Therefore, the only solution for $y$ is $y=9$. Substituting back,

$$
\frac{8 x^{2}}{2 x-1}=9, \quad 8 x^{2}-18 x+9=0
$$

The roots of this equation can be calculated using the quadratic formula:

$$
x_{1}=\frac{3}{4} \quad \text { and } \quad x_{2}=\frac{3}{2}
$$

which, as we can verify by substitution, are indeed solutions to the equation.

II. solution. Let $y=\frac{x}{2 x-1}$. Rearranging this,

$$
x=2 x y-y, \quad 2 x y=x+y .
$$

We see that if $x+y=a$, then $2 x y=a$. With this new notation, the left side of the original equation is

$$
x^{3}+y^{3}=(x+y)\left(x^{2}-x y+y^{2}\right)=a\left(a^{2}-3 x y\right)=a\left(a^{2}-\frac{3}{2} a\right) .
$$

Substituting, multiplying by 64, and rearranging, we get the integer-coefficient cubic equation

$$
64 a^{3}-96 a^{2}-243=0
$$

As already known from the first solution, if the equation has a rational root, then its numerator is a divisor of 243, and its denominator is a divisor of 64. From this, it can be seen after a few attempts that $a=\frac{9}{4}$ is a solution to the equation, and the factor $(4 a-9)$ can be factored out of the equation:

$$
(4 a-9)\left(16 a^{2}+12 a+27\right)=0
$$

The discriminant of the quadratic equation $16 a^{2}+12 a+27=0$ is negative, so it has no real roots. Therefore, the only real root of the cubic equation is $a=\frac{9}{4}$. From this,

$$
2 x y=x\left(\frac{x}{2 x-1}\right)=\frac{9}{4}, \quad 8 x^{2}-18 x+9=0, \quad x_{1}=\frac{3}{2} \quad \text { and } x_{2}=\frac{3}{4} \text {. }
$$

It can be verified by substitution that both fractions are indeed solutions to the original equation.