Solve the following equation

$$
\sqrt[3]{2 x+1}+\sqrt[3]{4-x}=-\sqrt[3]{3}
$$
I. Solution. If we raise both sides of the original equation to the third power and transform the left side according to the identity

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

we arrive at the equation

$$
3 \sqrt[3]{2 x+1} \sqrt[3]{4-x}(\sqrt[3]{2 x+1}+\sqrt[3]{4-x})=-(x+8)
$$

after rearrangement. Here, according to (1), the expression in the parentheses on the left side is $-\sqrt[3]{3}$, and thus

$$
3 \sqrt[3]{3(2 x+1)(4-x)}=x+8
$$

Raising both sides to the third power again and rearranging, we obtain the cubic equation

$$
x^{3}+186 x^{2}-375 x+188=0
$$

The sum of the coefficients on the left side is zero, so the equation has the root 1. After factoring out the $(x-1)$ factor, the resulting quadratic polynomial can be factored, yielding the equation

$$
(x-1)^{2}(x+188)=0
$$

This equation has two roots, 1 and -188. The 1 is not a solution to the original equation, since if $x=1$, the left side is positive, while the right side is negative. The -188 is a solution to the (1) equation.

Remark. The question arises as to why the "false root" appears. Two numbers are exactly equal if and only if their cubes are, so raising the equations to the third power (unlike squaring) is an equivalent transformation. Therefore, (1) and (2) are equivalent equations, just as (3) and (4) are. The false root could only have appeared when we substituted the original (1) relationship back into the (2) equation: this is not necessarily an equivalent transformation. Consider in general: if we solve the equation obtained after cubing and substituting $u(x)+v(x)=c$ into

$$
u(x)+v(x)-c=0
$$

we get

$$
u^{3}(x)+v^{3}(x)+3 u(x) \cdot v(x) \cdot c-c^{3}=0
$$

From the identity

$$
u^{3}+v^{3}-c^{3}+3 u v c=\frac{1}{2}(u+v-c)\left[(u-v)^{2}+(u+c)^{2}+(v+c)^{2}\right]
$$

it is clear that the second factor can also be zero when the first - which is the left side of the original (1) equation after rearrangement - is not. For this, there must exist an $x_{0}$ such that $u\left(x_{0}\right)=v\left(x_{0}\right) \neq 0$, and the constant $c$ on the right side of the equation must be the negative of this common value.

This is exactly what happened in the (1) equation, where

$$
u(x)=\sqrt[3]{2 x+1}, \quad v(x)=\sqrt[3]{4-x} \quad \text { and } \quad c=-\sqrt[3]{3}
$$

Here, for $x_{0}=1$, indeed $u\left(x_{0}\right)=v\left(x_{0}\right)=-c \neq 0$, so if $x_{0}=1$, in (6) the second factor is zero, while the first is not. It can be seen that for any other $c$ value different from $-\sqrt[3]{3}$, the substitution method leads to an equation equivalent to the original.

The following solution avoids the "critical" substitution.

II. Solution. By introducing the unknowns $u=\sqrt[3]{2 x+1}$ and $v=\sqrt[3]{4-x}$, we obtain the system of equations

$$
\begin{gathered}
u+v=-\sqrt[3]{3} \\
u^{3}+2 v^{3}=9
\end{gathered}
$$

From the first equation, $u=-(v+\sqrt[3]{3})$. Cubing and substituting the obtained form of $u^{3}$ into the second equation, we get the equation

$$
v^{3}-3 \sqrt[3]{3} v^{2}-3 \sqrt[3]{9} v-4 \sqrt[3]{27}=0
$$

The left side of the equation can be written as

$$
v^{3}-(\sqrt{3})^{3}-3 \sqrt[3]{3}\left(v^{2}+\sqrt[3]{3} v+(\sqrt[3]{3})^{2}\right)
$$

Factoring the difference of two cubes:

$$
v^{3}-(\sqrt[3]{3})^{3}=(v-\sqrt[3]{3})\left(v^{2}+\sqrt[3]{3} v+(\sqrt[3]{3})^{2}\right)
$$

The identical positive $\left(v^{2}+\sqrt[3]{3} v+(\sqrt[3]{3})^{2}\right)$ factor can be factored out, and we finally get the equation

$$
\left(v^{2}+\sqrt[3]{3} v+(\sqrt[3]{3})^{2}\right) \cdot(v-\sqrt[3]{3}-3 \sqrt[3]{3})=0
$$

from which

$$
v=\sqrt[3]{4-x}=4 \sqrt[3]{3}
$$

and thus $x=-188$.

Since we performed equivalent transformations here, -188 is also a solution to the original (1) equation.