Solve the equation $(9-3 x) \cdot 3^{x}-(x-2)\left(x^{2}-5 x+6\right)=0$ in the set of real numbers.
Solution. The quadratic expression $x^{2}-5 x+6$ can be factored into $(x-2)(x-3)$. Substituting this into the equation and factoring out $(-3)$ from $(9-3 x)$, we get

$$
-3 \cdot(x-3) \cdot 3^{x}-(x-2)(x-2)(x-3)=0
$$

Further simplifying this, we arrive at the equation

$$
(x-3)\left(-3^{x+1}-(x-2)^{2}\right)=0
$$

From here, $x=3$, or $3^{x+1}=-(x-2)^{2}$, which is never possible because the powers of 3 are positive numbers.

Therefore, the only solution to the equation is $x=3$, and this is indeed a solution, as we have only performed identical transformations.