6.175. $\sqrt[3]{x-1}+\sqrt[3]{x-2}-\sqrt[3]{2 x-3}=0$.
## Solution.

From the condition $\sqrt[3]{x-1}+\sqrt[3]{x-2}=\sqrt[3]{2 x-3}$, raising both sides to the cube, we have

$$
\begin{aligned}
& x-1+3 \sqrt[3]{(x-1)^{2}(x-2)}+3 \sqrt[3]{(x-1)(x-2)^{2}}+x-2=2 x-3 \Leftrightarrow \\
& \Leftrightarrow 3 \sqrt[3]{(x-1)(x-2)}(\sqrt[3]{(x-1)}+\sqrt[3]{(x-2)})=0
\end{aligned}
$$

Since $\sqrt[3]{x-1}+\sqrt[3]{x-2}=\sqrt[3]{2 x-3}$, the equation takes the form

$$
3 \sqrt[3]{(x-1)(x-2)(2 x-3)}=0 \Leftrightarrow(x-1)(x-2)(2 x-3)=0
$$

from which $x_{1}=1, x_{2}=2, x_{3}=\frac{3}{2}$.

Answer: $x_{1}=1, x_{2}=2, x_{3}=\frac{3}{2}$.