## Task B-2.2.

Solve the equation: $\frac{1}{\sqrt[3]{x+1}+\sqrt[3]{2 x-1}}-\frac{1}{\sqrt[3]{12 x}}=0$.
## Solution.

The conditions for the equation to have a solution are as follows: $\sqrt[3]{x+1}+\sqrt[3]{2 x-1} \neq 0$ and $12 x \neq 0$, i.e., $x \neq 0$. From the first condition, $\sqrt[3]{x+1} \neq -\sqrt[3]{2 x-1}$, and by cubing, we get $x+1 \neq -2 x+1$, hence $x \neq 0$. If we multiply the given equation by the common denominator $(\sqrt[3]{x+1}+\sqrt[3]{2 x-1}) \cdot \sqrt[3]{12 x}$, we obtain

$$
\sqrt[3]{x+1}+\sqrt[3]{2 x-1}=\sqrt[3]{12 x}
$$

By cubing, we get

$$
\begin{gathered}
x+1+3 \cdot \sqrt[3]{(x+1)^{2}} \cdot \sqrt[3]{2 x-1}+3 \cdot \sqrt[3]{x+1} \cdot \sqrt[3]{(2 x-1)^{2}}+2 x-1=12 x \\
3 \cdot \sqrt[3]{x+1} \cdot \sqrt[3]{2 x-1} \cdot(\sqrt[3]{x+1}+\sqrt[3]{2 x-1})=9 x \\
\sqrt[3]{x+1} \cdot \sqrt[3]{2 x-1} \cdot(\sqrt[3]{x+1}+\sqrt[3]{2 x-1})=3 x
\end{gathered}
$$

Notice that the expression in parentheses is equal to the left side of equation (*), so we can write

$$
\sqrt[3]{x+1} \cdot \sqrt[3]{2 x-1} \cdot \sqrt[3]{12 x}=3 x
$$

After cubing, we get

$$
(x+1)(2 x-1) 12 x=27 x^{3}
$$

Since $x \neq 0$, we divide the equation by $3 x$. Then we have

$$
4(x+1)(2 x-1)=9 x^{2}
$$

After simplification, we obtain the quadratic equation $x^{2}-4 x+4=0$, which has the solution $x=2$, which is also the solution to the original equation.